Contents / Calculus / Applications of Derivatives

Chapter 3

Applications of Derivatives

Extrema, the Mean Value Theorem, curve sketching, optimization, related rates, and Newton's method.

Introduction

The derivative was built to answer one question: how fast is $f$ changing at this instant? This chapter turns that single number into a toolkit. Where the derivative vanishes, a function may peak or bottom out; where it is positive, the graph climbs; where its own derivative is positive, the graph bends upward. Read together, these signs let you sketch a curve you have never seen, find the cheapest can that holds a liter, or tell how fast a shadow lengthens as a man walks away from a lamp.

Everything here rests on one theorem, the Mean Value Theorem. It is the bridge between the derivative, which is defined at a point, and the behavior of a function across an interval. Every test in this chapter — increasing and decreasing, the first and second derivative tests, the concavity test — is a corollary of it, and each is proved here.

The second half of the chapter is about translation: turning a sentence about fences, ladders or profits into a function of one variable, and reading the answer back. That translation step is where most errors live, so the worked problems show every line of it.

3.1Maximum and minimum values

Some of the most important applications of calculus are optimization problems: find the largest, the smallest, the cheapest, the fastest. The first job is to say precisely what a maximum is, and then to discover where one can hide.

Definition 3.1 (Absolute and local extrema). Let $c$ be a number in the domain $D$ of $f$ . Then $f(c)$ is the absolute maximum value of $f$ on $D$ if $f(c) \ge f(x)$ for all $x$ in $D$ , and the absolute minimum value if $f(c) \le f(x)$ for all $x$ in $D$ .

The number $f(c)$ is a local maximum value of $f$ if $f(c) \ge f(x)$ for all $x$ in some open interval containing $c$ , and a local minimum value if $f(c) \le f(x)$ on some open interval containing $c$ . Maximum and minimum values are called extreme values, or extrema.

An absolute extremum is the largest or smallest value the function ever takes; a local extremum only has to win against its immediate neighbors. Every absolute extremum attained at an interior point is also a local one, but not the other way around. The function $f(x) = \cos x$ has absolute maximum $1$ , reached infinitely often, at every $x = 2n\pi$ ; each of these is a local maximum as well. The function $f(x) = x^{2}$ has absolute (and local) minimum $0$ at $x = 0$ and no maximum of any kind. The function $f(x) = x^{3}$ has no extrema at all: it is increasing everywhere, so every value is beaten by the next.

The domain matters. On the whole real line $x^{2}$ has no maximum, but on $[-1, 2]$ its maximum is $4$ , at the endpoint $x = 2$ . Because a local extremum is compared with points on both sides, an endpoint can carry an absolute extremum but never a local one. That convention is what makes the closed interval method below work: interior extrema are local, and local extrema are caught by the derivative, while endpoints are simply checked by hand.

Theorem 3.2 (Extreme Value Theorem). If $f$ is continuous on a closed interval $[a, b]$ , then $f$ attains an absolute maximum value $f(c)$ and an absolute minimum value $f(d)$ at some numbers $c$ and $d$ in $[a, b]$ .

The theorem is intuitively clear: a continuous graph over a closed interval is an unbroken curve with two endpoints, and such a curve has a highest point and a lowest point. The proof belongs to real analysis, because it depends on the completeness of the real numbers — the fact that a bounded set of reals has a least upper bound — and that is why it is stated here without proof. What can be shown here is that both hypotheses are needed.

Drop closedness and the theorem fails: $f(x) = x$ on the open interval $(0, 1)$ takes values as close to $1$ as you like but never reaches $1$ , so it has no maximum, and no minimum either. Drop continuity and it fails again: the function on $[0, 2]$ equal to $x$ for $x < 1$ and to $0$ for $x \ge 1$ climbs toward $1$ , jumps down, and never attains a largest value. The theorem promises that extrema exist; it says nothing about where. Finding them is the work of the next result.

Theorem 3.3 (Fermat's Theorem). If $f$ has a local maximum or minimum at $c$ , and $f'(c)$ exists, then $f'(c) = 0$ .

Proof. Suppose $f$ has a local maximum at $c$ , so $f(x) \le f(c)$ for every $x$ close enough to $c$ , and therefore $f(c+h) - f(c) \le 0$ for every $h$ small enough in absolute value. Because $f'(c)$ exists, both one-sided limits of the difference quotient exist and equal $f'(c)$ .

For $h > 0$ the quotient is a non-positive number over a positive one,

\frac{f(c+h)-f(c)}{h} \le 0,

so letting $h \to 0^+$ gives $f'(c) \le 0$ .

For $h < 0$ the same numerator is divided by a negative number, so

\frac{f(c+h)-f(c)}{h} \ge 0,

and letting $h \to 0^-$ gives $f'(c) \ge 0$ .

The only number that is both $\le 0$ and $\ge 0$ is $0$ , so $f'(c) = 0$ . The case of a local minimum is identical with the inequalities reversed, or follows by applying this case to $-f$ .∎

Fermat's Theorem says that at a local extremum the tangent line is horizontal — provided there is a tangent line. Two cautions keep it honest. First, the converse is false: $f(x) = x^{3}$ has $f'(0) = 0$ , yet $0$ is not a local extremum, since $x^{3}$ is negative just to the left and positive just to the right. Second, a local extremum need not have a derivative at all: $f(x) = |x|$ has its absolute and local minimum at $0$ , where $f'(0)$ does not exist. So $f'(c) = 0$ is a place to look for extrema, not a guarantee, and places where $f'$ fails to exist must be searched too. That pair of conditions deserves a name.

Definition 3.4 (Critical number). A critical number of a function $f$ is a number $c$ in the domain of $f$ such that either $f'(c) = 0$ or $f'(c)$ does not exist.

Proposition 3.5 (Extrema occur at critical numbers). If $f$ has a local maximum or minimum at $c$ , then $c$ is a critical number of $f$ .

Proof. Either $f'(c)$ exists, in which case Fermat's Theorem gives $f'(c) = 0$ , or it does not exist. Both cases are covered by the definition of a critical number.∎

Combine this proposition with the Extreme Value Theorem and the earlier remark about endpoints. On a closed interval, the absolute maximum of a continuous function exists; it is attained either at an endpoint or at an interior point, and an interior absolute maximum is a local maximum, hence at a critical number. There is nowhere else for it to be.

Method 3.6 (The Closed Interval Method). To find the absolute maximum and minimum values of a continuous function $f$ on a closed interval $[a, b]$ :

Find the critical numbers of $f$ in the open interval $(a, b)$ : solve $f'(x) = 0$ and locate any points where $f'$ is undefined.
Evaluate $f$ at each of those critical numbers.
Evaluate $f$ at the endpoints $a$ and $b$ .
The largest of the values from steps 2 and 3 is the absolute maximum; the smallest is the absolute minimum.

Intuition. Walk a hiking trail from trailhead A to trailhead B. Somewhere along it there is a highest point and a lowest point — that is the Extreme Value Theorem. Where could the highest point be? Either at one of the trailheads, or at a summit in the middle. At a smooth summit the ground is momentarily level (that is Fermat: the slope is zero), and at a jagged rocky peak the slope is undefined. So the candidates are: the two trailheads, every level spot, and every sharp spot. Measure the altitude at each candidate and pick the largest. That is the whole closed interval method.

Example 3.7 (Finding critical numbers). Find the critical numbers of $f(x) = x^{3/5}(4 - x)$ .

Solution.

Expand first so the differentiation is a pair of power rules: $f(x) = 4x^{3/5} - x^{8/5}$ .
Differentiate:

f'(x) = \frac{12}{5}x^{-2/5} - \frac{8}{5}x^{3/5} = \frac{4}{5}x^{-2/5}\left(3 - 2x\right) = \frac{4(3-2x)}{5x^{2/5}}.

$f'(x) = 0$ when the numerator vanishes: $3 - 2x = 0$ , so $x = \frac{3}{2}$ .
$f'(x)$ does not exist when the denominator vanishes, at $x = 0$ . Since $f(0) = 0$ is defined, $0$ is in the domain and counts.
The critical numbers are $x = 0$ and $x = \frac{3}{2}$ . Sanity check: near $x = 0$ the graph has a vertical tangent (the factor $x^{3/5}$ rises steeply), which is exactly the kind of point Fermat's Theorem cannot see.

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Example 3.8 (The closed interval method for a cubic). Find the absolute maximum and minimum values of $f(x) = x^{3} - 3x^{2} + 1$ on $\left[-\tfrac12, 4\right]$ .

Solution.

$f$ is a polynomial, so it is continuous and the Extreme Value Theorem applies.
$f'(x) = 3x^{2} - 6x = 3x(x - 2)$ , which exists everywhere and vanishes at $x = 0$ and $x = 2$ . Both lie in $\left(-\tfrac12, 4\right)$ .
Values at the critical numbers: $f(0) = 1$ and $f(2) = 8 - 12 + 1 = -3$ .
Values at the endpoints: $f\!\left(-\tfrac12\right) = -\tfrac18 - \tfrac34 + 1 = \tfrac18$ and $f(4) = 64 - 48 + 1 = 17$ .
Comparing $1$ , $-3$ , $\tfrac18$ , $17$ : the absolute maximum is $f(4) = 17$ and the absolute minimum is $f(2) = -3$ .

The maximum sits at an endpoint and the minimum at an interior critical number. Skip either kind of candidate and you get the wrong answer.□

Example 3.9 (The closed interval method with trigonometry). Find the absolute extrema of $f(\theta) = 2\cos\theta + \sin 2\theta$ on $\left[0, \tfrac{\pi}{2}\right]$ .

Solution.

$f'(\theta) = -2\sin\theta + 2\cos 2\theta$ . Use $\cos 2\theta = 1 - 2\sin^{2}\theta$ to write everything in terms of $\sin\theta$ :

f'(\theta) = -2\sin\theta + 2 - 4\sin^2\theta = -2\left(2\sin^2\theta + \sin\theta - 1\right) = -2(2\sin\theta - 1)(\sin\theta + 1).

On $\left(0, \tfrac{\pi}{2}\right)$ we have $\sin\theta + 1 > 0$ , so $f'(\theta) = 0$ only when $\sin\theta = \tfrac12$ , that is $\theta = \tfrac{\pi}{6}$ .
$f\!\left(\tfrac{\pi}{6}\right) = 2\cdot\tfrac{\sqrt3}{2} + \sin\tfrac{\pi}{3} = \sqrt3 + \tfrac{\sqrt3}{2} = \tfrac{3\sqrt3}{2} \approx 2.60$ .
Endpoints: $f(0) = 2 + 0 = 2$ and $f\!\left(\tfrac{\pi}{2}\right) = 0 + \sin\pi = 0$ .
Absolute maximum $\tfrac{3\sqrt3}{2}$ at $\theta = \tfrac{\pi}{6}$ ; absolute minimum $0$ at $\theta = \tfrac{\pi}{2}$ .

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Pitfall. The closed interval method needs a closed interval and a continuous function. On $(0, \infty)$ the function $x + \tfrac{4}{x}$ has a minimum but no maximum, and no table of endpoint values will tell you that; open-interval problems are handled by the first derivative test for absolute extrema in the optimization section. And a critical number outside the interval must be discarded: for $x^{3} - 3x$ on $[0, 2]$ , the critical number $x = -1$ plays no part.

Exercises for §3.1 →

3.2The Mean Value Theorem

The derivative is defined at a point. Almost everything we want to know — is $f$ increasing on this interval, how far can $f$ move if its derivative is small — concerns an interval. The Mean Value Theorem is the link, and it is proved through a special case that is worth stating on its own.

Theorem 3.10 (Rolle's Theorem). Let $f$ satisfy three hypotheses: $f$ is continuous on the closed interval $[a, b]$ ; $f$ is differentiable on the open interval $(a, b)$ ; and $f(a) = f(b)$ . Then there is a number $c$ in $(a, b)$ with $f'(c) = 0$ .

Proof. There are three cases.

If $f$ is constant on $[a, b]$ , then $f'(x) = 0$ for every $x$ in $(a, b)$ and any $c$ will do.

If $f(x) > f(a)$ for some $x$ in $(a, b)$ , then by the Extreme Value Theorem (continuity on a closed interval) $f$ attains an absolute maximum at some $c$ in $[a, b]$ . That maximum value exceeds $f(a) = f(b)$ , so $c$ is not an endpoint: it lies in the open interval $(a, b)$ , where the absolute maximum is also a local maximum. Since $f$ is differentiable at $c$ , Fermat's Theorem gives $f'(c) = 0$ .

If $f(x) < f(a)$ for some $x$ in $(a, b)$ , the same argument with the absolute minimum gives a $c$ in $(a, b)$ with $f'(c) = 0$ .∎

Each hypothesis earns its place. Differentiability fails for $f(x) = |x|$ on $[-1, 1]$ : it is continuous, $f(-1) = f(1)$ , but the derivative is never zero — the only candidate is the corner at $0$ , where there is no derivative. Continuity fails for the function equal to $x$ on $[0, 1)$ and $0$ at $x = 1$ : the endpoint values agree, the derivative is $1$ everywhere it exists, and no horizontal tangent appears. Equal endpoint values fail for $f(x) = x$ on $[0, 1]$ , whose derivative is identically $1$ .

Rolle's Theorem says that a smooth curve which returns to its starting height must be level somewhere in between. Tilt the picture and you get the general statement.

Theorem 3.11 (The Mean Value Theorem). Let $f$ be continuous on $[a, b]$ and differentiable on $(a, b)$ . Then there is a number $c$ in $(a, b)$ such that

f'(c) = \frac{f(b) - f(a)}{b - a},

or equivalently $f(b) - f(a) = f'(c)(b - a)$ .

Proof. The right-hand side is the slope of the secant line through $(a, f(a))$ and $(b, f(b))$ . Subtract that line from $f$ : define

h(x) = f(x) - f(a) - \frac{f(b) - f(a)}{b - a}(x - a).

Then $h$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , because $f$ is and the subtracted part is a polynomial. At the endpoints, $h(a) = f(a) - f(a) - 0 = 0$ and

h(b) = f(b) - f(a) - \frac{f(b) - f(a)}{b - a}(b - a) = 0.

So $h$ satisfies the hypotheses of Rolle's Theorem, and there is a $c$ in $(a, b)$ with $h'(c) = 0$ . But

h'(x) = f'(x) - \frac{f(b) - f(a)}{b - a},

so $h'(c) = 0$ is precisely $f'(c) = \dfrac{f(b) - f(a)}{b - a}$ .∎

Geometrically: somewhere between $a$ and $b$ the tangent line is parallel to the secant line joining the endpoints. In the language of rates: over any interval, the instantaneous rate of change equals the average rate of change at least once. The theorem does not say where $c$ is, and there may be several such $c$ ; it only says one exists. That is enough for everything that follows.

Intuition. You drive $120$ miles in $2$ hours, an average of $60$ mph. At some instant the speedometer read exactly $60$ . It cannot have read below $60$ the whole way, or you would have covered fewer than $120$ miles; it cannot have read above $60$ the whole way either. A speedometer needle moves continuously, so on its way from below $60$ to above $60$ (or the reverse) it passes through $60$ . Police forces use exactly this: two cameras $10$ miles apart that clock you passing $8$ minutes apart know, by the Mean Value Theorem, that you touched $75$ mph somewhere in between.

Example 3.12 (Finding the number guaranteed by the theorem). Find every number $c$ that satisfies the conclusion of the Mean Value Theorem for $f(x) = x^{3} - x$ on $[0, 2]$ .

Solution.

$f$ is a polynomial, so it is continuous on $[0, 2]$ and differentiable on $(0, 2)$ ; the theorem applies.
The average rate of change is $\dfrac{f(2) - f(0)}{2 - 0} = \dfrac{6 - 0}{2} = 3$ .
$f'(x) = 3x^{2} - 1$ . Solve $3c^{2} - 1 = 3$ : $c^{2} = \tfrac43$ , so $c = \pm\tfrac{2}{\sqrt3}$ .
Only $c = \tfrac{2}{\sqrt3} \approx 1.15$ lies in $(0, 2)$ . The negative root is discarded.

Sanity check: the tangent at $x = 1.15$ has slope $3(1.33) - 1 = 3$ , matching the secant.□

Example 3.13 (Bounding a function by bounding its derivative). Suppose $f(0) = -3$ and $f'(x) \le 5$ for all $x$ . How large can $f(2)$ possibly be?

Solution.

$f$ is differentiable everywhere, hence continuous, so the Mean Value Theorem applies on $[0, 2]$ : there is a $c$ in $(0, 2)$ with $f(2) - f(0) = f'(c)\cdot 2$ .
Therefore $f(2) = -3 + 2f'(c) \le -3 + 2\cdot 5 = 7$ .
The largest possible value is $7$ , and it is attained by $f(x) = 5x - 3$ , so the bound is sharp.

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Example 3.14 (Counting roots with Rolle's Theorem). Show that the equation $x^{3} + x - 1 = 0$ has exactly one real root.

Solution.

Existence: $f(x) = x^{3} + x - 1$ is continuous with $f(0) = -1 < 0$ and $f(1) = 1 > 0$ , so by the Intermediate Value Theorem there is a root in $(0, 1)$ .
Uniqueness: suppose there were two roots $a < b$ . Then $f(a) = f(b) = 0$ , and Rolle's Theorem would give a $c$ in $(a, b)$ with $f'(c) = 0$ .
But $f'(x) = 3x^{2} + 1 \ge 1 > 0$ for every $x$ , so $f'$ is never zero. The supposition is impossible, and the root is unique.

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The Mean Value Theorem's first consequences look almost too obvious to need proof, yet without the theorem there is no way to prove them.

Corollary 3.15 (Zero derivative means constant). If $f'(x) = 0$ for all $x$ in an interval $(a, b)$ , then $f$ is constant on $(a, b)$ .

Proof. Take any two points $x_1 < x_2$ in $(a, b)$ . Since $f$ is differentiable on $(a, b)$ it is continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$ , so the Mean Value Theorem gives a $c$ with $f(x_2) - f(x_1) = f'(c)(x_2 - x_1) = 0$ . Thus $f(x_1) = f(x_2)$ : any two values of $f$ agree, which is what constant means.∎

Corollary 3.16 (Equal derivatives differ by a constant). If $f'(x) = g'(x)$ for all $x$ in an interval $(a, b)$ , then there is a constant $C$ with $f(x) = g(x) + C$ on $(a, b)$ .

Proof. Apply the previous corollary to $F = f - g$ , whose derivative $f' - g'$ is zero on $(a, b)$ . Then $F$ is constant, say $C$ , and $f = g + C$ .∎

This second corollary is the reason an antiderivative is determined only up to $+ C$ , which the integrals chapter will use on every page. It also proves identities. On $(-1, 1)$ ,

\frac{d}{dx}\left(\sin^{-1}x + \cos^{-1}x\right) = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0,

so the sum is constant there; evaluating at $x = 0$ gives the constant $\tfrac{\pi}{2}$ , and so $\sin^{-1}x + \cos^{-1}x = \tfrac{\pi}{2}$ for all $x$ in $(-1, 1)$ .

Pitfall. The word interval in the corollary is not decoration. The function $f(x) = x/|x|$ has $f'(x) = 0$ at every point of its domain, yet $f = -1$ on the left and $f = 1$ on the right. Its domain, $(-\infty, 0) \cup (0, \infty)$ , is not an interval, so the Mean Value Theorem cannot cross the gap at $0$ and the corollary does not apply.

Exercises for §3.2 →

3.3How derivatives determine the shape of a graph

The Mean Value Theorem converts information about $f'$ at every point into information about $f$ over an interval. The first payoff is the test that every sign chart in this chapter silently uses.

Theorem 3.17 (Increasing/Decreasing Test). Let $f$ be continuous on an interval $I$ and differentiable in its interior.

If $f'(x) > 0$ on the interior of $I$ , then $f$ is increasing on $I$ .

If $f'(x) < 0$ on the interior of $I$ , then $f$ is decreasing on $I$ .

Proof. Suppose $f'(x) > 0$ throughout, and take any $x_1 < x_2$ in $I$ . The Mean Value Theorem on $[x_1, x_2]$ gives a $c$ in $(x_1, x_2)$ with

f(x_2) - f(x_1) = f'(c)(x_2 - x_1).

Both factors on the right are positive — $f'(c) > 0$ by hypothesis and $x_2 - x_1 > 0$ by choice — so $f(x_2) - f(x_1) > 0$ , that is $f(x_1) < f(x_2)$ . Since $x_1 < x_2$ were arbitrary, $f$ is increasing. The decreasing case is identical with $f'(c) < 0$ .∎

Notice how little the proof needs and how much it gives. The hypothesis is about the derivative at single points; the conclusion compares values at points possibly far apart, and only the Mean Value Theorem can carry information across that gap. Notice also that the conclusion holds on the closed interval even though the hypothesis is only about the interior: $f(x) = x^{2}$ has $f'(0) = 0$ but is still increasing on $[0, \infty)$ .

To use the test, find where $f'$ can change sign. A continuous $f'$ changes sign only at its zeros, and $f'$ can also flip across a point where it fails to exist — so the critical numbers, together with points outside the domain, cut the line into intervals on which $f'$ keeps one sign. Pick a test value inside each interval, evaluate the sign of $f'$ there, and the sign holds across the whole interval. That table is called a sign chart.

Theorem 3.18 (First Derivative Test). Suppose $c$ is a critical number of a continuous function $f$ .

If $f'$ changes from positive to negative at $c$ , then $f$ has a local maximum at $c$ .

If $f'$ changes from negative to positive at $c$ , then $f$ has a local minimum at $c$ .

If $f'$ does not change sign at $c$ — positive on both sides, or negative on both sides — then $f$ has no local extremum at $c$ .

Proof. Suppose $f' > 0$ on an interval $(a, c)$ and $f' < 0$ on an interval $(c, b)$ . By the Increasing/Decreasing Test, $f$ is increasing on $(a, c]$ and decreasing on $[c, b)$ , using continuity at $c$ to include the endpoint in each. Increasing on the left gives $f(x) \le f(c)$ for $x$ in $(a, c]$ , and decreasing on the right gives $f(c) \ge f(x)$ for $x$ in $[c, b)$ . So $f(c) \ge f(x)$ on the whole interval $(a, b)$ , which is a local maximum. The second case is the same argument with the inequalities reversed. In the third case $f$ is increasing (or decreasing) on both sides of $c$ , hence on the whole interval by continuity, so values to the left of $c$ are smaller and values to the right are larger: neither a maximum nor a minimum.∎

The test is exhaustive and it never fails. It applies where $f'(c)$ does not exist — at the corner of $|x|$ , where $f'$ goes from $-1$ to $+1$ and the test correctly reports a minimum — and it applies where the second derivative test will later be silent.

Intuition. Walk left to right along the graph. $f' > 0$ means you are walking uphill, $f' < 0$ means downhill. A local maximum is the instant walking uphill turns into walking downhill: you just crossed a summit. A local minimum is downhill turning into uphill: the bottom of a valley. If you were going uphill before and after, you merely stepped over a level patch on the way up — no summit, no valley. That level patch is $x^{3}$ at the origin.

Example 3.19 (Increase, decrease, and local extrema). Find the intervals of increase and decrease and all local extrema of $f(x) = 3x^{4} - 4x^{3} - 12x^{2} + 5$ .

Solution.

$f'(x) = 12x^{3} - 12x^{2} - 24x = 12x(x^{2} - x - 2) = 12x(x - 2)(x + 1)$ .
$f'$ exists everywhere, and vanishes at $x = -1$ , $x = 0$ and $x = 2$ : three critical numbers cutting the line into four intervals.
Sign chart, using one test value per interval. On $(-\infty, -1)$ take $x = -2$ : $12(-2)(-4)(-1) = -96 < 0$ . On $(-1, 0)$ take $x = -\tfrac12$ : $12(-\tfrac12)(-\tfrac52)(\tfrac12) = 7.5 > 0$ . On $(0, 2)$ take $x = 1$ : $12(1)(-1)(2) = -24 < 0$ . On $(2, \infty)$ take $x = 3$ : $12(3)(1)(4) = 144 > 0$ .
So $f$ decreases on $(-\infty, -1)$ , increases on $(-1, 0)$ , decreases on $(0, 2)$ , increases on $(2, \infty)$ .
First Derivative Test: at $x = -1$ , $f'$ goes $-$ to $+$ , a local minimum, $f(-1) = 3 + 4 - 12 + 5 = 0$ . At $x = 0$ , $f'$ goes $+$ to $-$ , a local maximum, $f(0) = 5$ . At $x = 2$ , $f'$ goes $-$ to $+$ , a local minimum, $f(2) = 48 - 32 - 48 + 5 = -27$ .

Sanity check: the pattern down-up-down-up must alternate min, max, min, and it does.□

Example 3.20 (A critical number where the derivative does not exist). Find the local extrema of $f(x) = x^{2/3}(x - 5)$ .

Solution.

Write $f(x) = x^{5/3} - 5x^{2/3}$ , so

f'(x) = \frac53 x^{2/3} - \frac{10}{3}x^{-1/3} = \frac{5(x - 2)}{3x^{1/3}}.

$f'(x) = 0$ at $x = 2$ ; $f'$ does not exist at $x = 0$ , which is in the domain. Both are critical numbers.
Signs: for $x < 0$ both $x - 2$ and $x^{1/3}$ are negative, so $f' > 0$ . For $0 < x < 2$ the numerator is negative and the denominator positive, so $f' < 0$ . For $x > 2$ both are positive, so $f' > 0$ .
At $x = 0$ : $+$ to $-$ , a local maximum, $f(0) = 0$ . At $x = 2$ : $-$ to $+$ , a local minimum, $f(2) = 2^{2/3}(-3) = -3\sqrt[3]{4} \approx -4.76$ .

The local maximum at $x = 0$ is a cusp: $f'$ blows up to $+\infty$ on one side and $-\infty$ on the other. Fermat's Theorem could not have found it, but the First Derivative Test has no trouble.□

Pitfall. A sign chart is built from the critical numbers and from the numbers where $f$ itself is undefined. For $f(x) = \tfrac1x$ the derivative $-\tfrac{1}{x^{2}}$ is never zero, so there are no critical numbers, yet the sign chart still needs a break at $x = 0$ — and $0$ is not a critical number, because it is not in the domain. It follows that $f$ is decreasing on $(-\infty, 0)$ and on $(0, \infty)$ , but not on their union: $f(-1) = -1$ is less than $f(1) = 1$ .

Exercises for §3.3 →

3.4Concavity and inflection points

Increasing or decreasing describes where the graph is going. Concavity describes how it bends on the way, and it is what separates a curve that flattens out from one that accelerates away.

Definition 3.21 (Concavity). If the graph of $f$ lies above all of its tangent lines on an interval $I$ , then $f$ is called concave upward on $I$ . If the graph lies below all of its tangents on $I$ , $f$ is concave downward on $I$ .

An equivalent and often handier description: $f$ is concave upward exactly when $f'$ is an increasing function, and concave downward exactly when $f'$ is decreasing. Slopes that grow as you move right produce a curve that turns leftward, opening up like a bowl; slopes that shrink produce an arch. Since " $f'$ is increasing" is decided by the sign of $(f')' = f''$ , the Increasing/Decreasing Test applied to $f'$ gives the working test at once.

Theorem 3.22 (Concavity Test). If $f''(x) > 0$ for all $x$ in an interval $I$ , then $f$ is concave upward on $I$ . If $f''(x) < 0$ for all $x$ in $I$ , then $f$ is concave downward on $I$ .

Proof. Suppose $f'' > 0$ on $I$ and fix $a$ in $I$ ; we show the graph lies above the tangent at $a$ , whose equation is $y = f(a) + f'(a)(x - a)$ . Take $x > a$ in $I$ . The Mean Value Theorem on $[a, x]$ gives a $c$ in $(a, x)$ with $f(x) - f(a) = f'(c)(x - a)$ . Because $f'' > 0$ , the Increasing/Decreasing Test applied to $f'$ says $f'$ is increasing, and $c > a$ , so $f'(c) > f'(a)$ . Multiplying by the positive number $x - a$ ,

f(x) - f(a) = f'(c)(x-a) > f'(a)(x-a),

which rearranges to $f(x) > f(a) + f'(a)(x-a)$ : the graph is above the tangent. For $x < a$ the same Mean Value Theorem gives a $c$ in $(x, a)$ with $f'(c) < f'(a)$ , and multiplying by the negative number $x - a$ reverses the inequality to the same conclusion. The concave downward case follows by applying this to $-f$ .∎

Definition 3.23 (Inflection point). A point $P$ on a curve $y = f(x)$ is called an inflection point if $f$ is continuous there and the curve changes from concave upward to concave downward, or from concave downward to concave upward, at $P$ .

To hunt for inflection points, list the numbers where $f''(x) = 0$ or $f''$ fails to exist, then check that $f''$ actually changes sign there. Both halves matter. For $f(x) = x^{4}$ we have $f''(x) = 12x^{2}$ , which is zero at $x = 0$ but positive on both sides: the curve is concave up throughout and $(0,0)$ is not an inflection point. For $f(x) = x^{1/3}$ the second derivative $-\tfrac29 x^{-5/3}$ never vanishes but does not exist at $x = 0$ , and it changes sign there, so the origin is an inflection point — with a vertical tangent.

Intuition. Drive along the curve as if it were a road. Concave up means you are turning the steering wheel to the left; concave down means turning right. An inflection point is the instant the wheel passes through center as you switch from one turn to the other.

Concavity is also the difference between good news and bad news about a trend. If the number of new cases in an epidemic is still rising ( $f' > 0$ ) but the curve has turned concave down ( $f'' < 0$ ), the rise is slowing: the inflection point is the day the outbreak passed its worst rate of growth, even though the total is still climbing.

Example 3.24 (Concavity and inflection points of a quartic). Discuss the concavity of $f(x) = x^{4} - 4x^{3}$ and find its inflection points.

Solution.

$f'(x) = 4x^{3} - 12x^{2}$ and $f''(x) = 12x^{2} - 24x = 12x(x - 2)$ .
$f'' = 0$ at $x = 0$ and $x = 2$ ; $f''$ exists everywhere.
Signs: on $(-\infty, 0)$ take $x = -1$ , $f'' = 12(-1)(-3) = 36 > 0$ , concave up. On $(0, 2)$ take $x = 1$ , $f'' = 12(1)(-1) = -12 < 0$ , concave down. On $(2, \infty)$ take $x = 3$ , $f'' = 12(3)(1) = 36 > 0$ , concave up.
$f''$ changes sign at both numbers, so both give inflection points: $(0, 0)$ and $(2, -16)$ .

Note that $x = 0$ is a critical number too — $f'(0) = 0$ — but the graph has neither a maximum nor a minimum there, only a flat inflection.□

Example 3.25 (Concavity of an exponential model). On what intervals is $f(x) = xe^{-x}$ concave upward, and where is its inflection point?

Solution.

Product rule: $f'(x) = e^{-x} - xe^{-x} = (1 - x)e^{-x}$ .
Again: $f''(x) = -e^{-x} - (1-x)e^{-x} = (x - 2)e^{-x}$ .
Since $e^{-x} > 0$ always, the sign of $f''$ is the sign of $x - 2$ . So $f$ is concave downward on $(-\infty, 2)$ and concave upward on $(2, \infty)$ .
The concavity changes at $x = 2$ , so the inflection point is $\left(2, 2e^{-2}\right) \approx (2, 0.27)$ .

Sanity check against the first derivative: $f' = 0$ at $x = 1$ , a local maximum, and the curve is concave down there, as an arch should be.□

Exercises for §3.4 →

3.5The second derivative test

At a critical number with $f'(c) = 0$ , the concavity near $c$ decides the question immediately, with no sign chart to build.

Theorem 3.26 (Second Derivative Test). Suppose $f''$ is continuous near $c$ .

If $f'(c) = 0$ and $f''(c) > 0$ , then $f$ has a local minimum at $c$ .

If $f'(c) = 0$ and $f''(c) < 0$ , then $f$ has a local maximum at $c$ .

Proof. Suppose $f'(c) = 0$ and $f''(c) > 0$ . Because $f''$ is continuous at $c$ and positive there, $f'' > 0$ on some open interval $I$ containing $c$ . By the Concavity Test, $f$ is concave upward on $I$ , so on $I$ the graph lies above the tangent line at $c$ . That tangent is the horizontal line $y = f(c)$ , because $f'(c) = 0$ . Hence $f(x) > f(c)$ for every $x \ne c$ in $I$ : a local minimum. The other case follows by applying this to $-f$ .∎

The test is quick and it is the one to reach for when $f''$ is easy to compute — polynomials, exponentials, most textbook optimization problems. It has two blind spots, and both are worth knowing precisely.

It says nothing when $f''(c) = 0$ . All three behaviors occur: $f(x) = x^{4}$ has $f''(0) = 0$ and a minimum; $f(x) = -x^{4}$ has $f''(0) = 0$ and a maximum; $f(x) = x^{3}$ has $f''(0) = 0$ and neither. When the test is inconclusive, fall back on the First Derivative Test, which always decides.

It also says nothing when $f'(c)$ fails to exist, since it assumes $f'(c) = 0$ outright. The cusp of $x^{2/3}(x-5)$ at the origin is invisible to it. The First Derivative Test handles that case too.

Intuition. At a critical number the graph is momentarily flat, and the second derivative says which way it is about to bend. Positive means the curve smiles around you — you are at the bottom of a bowl, a minimum. Negative means it frowns — you are on top of an arch, a maximum. Zero means the bend is too gentle to read at this order and you must look at what happens on each side instead.

Example 3.27 (Second derivative test on a cubic). Use the Second Derivative Test to classify the critical numbers of $f(x) = 2x^{3} - 3x^{2} - 36x$ .

Solution.

$f'(x) = 6x^{2} - 6x - 36 = 6(x^{2} - x - 6) = 6(x - 3)(x + 2)$ , so the critical numbers are $x = 3$ and $x = -2$ .
$f''(x) = 12x - 6$ .
$f''(3) = 30 > 0$ : local minimum, value $f(3) = 54 - 27 - 108 = -81$ .
$f''(-2) = -30 < 0$ : local maximum, value $f(-2) = -16 - 12 + 72 = 44$ .

Sanity check: a cubic with positive leading coefficient rises, dips and rises, so the maximum must come before the minimum — and $-2 < 3$ .□

Example 3.28 (When the test is inconclusive). Classify the critical numbers of $f(x) = x^{4} - 4x^{3}$ .

Solution.

$f'(x) = 4x^{3} - 12x^{2} = 4x^{2}(x - 3)$ , so the critical numbers are $x = 0$ and $x = 3$ .
$f''(x) = 12x^{2} - 24x$ . At $x = 3$ : $f''(3) = 108 - 72 = 36 > 0$ , a local minimum with $f(3) = 81 - 108 = -27$ .
At $x = 0$ : $f''(0) = 0$ , so the Second Derivative Test is silent.
Use the First Derivative Test instead. Near $x = 0$ the factor $4x^{2}$ is positive and $x - 3$ is negative on both sides, so $f' < 0$ on $(-1, 0)$ and on $(0, 1)$ . No sign change, so $x = 0$ is not a local extremum — it is the flat inflection point found earlier.

□

Pitfall. Confirming a critical number is a local minimum does not make it the absolute minimum, and a problem that asks for the smallest value on an interval is not finished until the endpoints have been checked or the function's behavior at the ends of an open interval has been argued. The optimization section gives the tool for the open-interval case.

Exercises for §3.5 →

3.6Curve sketching

Calculators plot points; you are about to do something they cannot, which is to know in advance exactly what the graph must look like and why. Every feature worth marking comes from one of the tests already proved, and the checklist below simply runs through them in an order that never doubles back.

Method 3.29 (Guidelines for sketching a curve).

Domain. Find the set of $x$ for which $f(x)$ is defined. Every later step is restricted to it.
Intercepts. The $y$ -intercept is $f(0)$ ; the $x$ -intercepts solve $f(x) = 0$ . Skip them if the equation is hard.
Symmetry. If $f(-x) = f(x)$ the curve is even, symmetric about the $y$ -axis; if $f(-x) = -f(x)$ it is odd, symmetric about the origin. Either halves the work. Check also for periodicity.
Asymptotes. Horizontal: compute $\lim_{x\to\pm\infty} f(x)$ ; a finite limit $L$ gives the asymptote $y = L$ . Vertical: at each $a$ excluded from the domain, compute the one-sided limits; an infinite one gives the asymptote $x = a$ . Slant: if $\lim_{x\to\pm\infty}[f(x) - (mx+b)] = 0$ , the line $y = mx + b$ is a slant asymptote — for a rational function whose numerator exceeds the denominator in degree by exactly one, long division produces it.
Intervals of increase and decrease. Compute $f'$ , find the critical numbers, build the sign chart.
Local extrema. Apply the First Derivative Test (or the Second, where $f''$ is cheap) at each critical number.
Concavity and inflection points. Compute $f''$ , find where it is zero or undefined, build its sign chart, and keep the places where the sign actually changes.
Sketch. Draw the asymptotes as dashed lines, plot the intercepts, extrema and inflection points, then join them respecting the direction from step 5 and the bend from step 7.

Intuition. The checklist is a police sketch built from witness statements. Each step is a witness: the domain says where the suspect can be, the asymptotes say what walls it hugs, the first derivative says it climbs here and falls there, the second says it bends this way. Any one statement leaves a thousand possible faces. All of them together leave essentially one.

Example 3.30 (A rational function with a slant asymptote). Sketch the curve $y = \dfrac{x^{2}}{x - 1}$ .

Solution.

Domain: all $x \ne 1$ .
Intercepts: $f(0) = 0$ , and $x^{2} = 0$ gives $x = 0$ . The curve meets the axes only at the origin.
Symmetry: $f(-x) = \frac{x^{2}}{-x-1}$ , which is neither $f(x)$ nor $-f(x)$ . None.
Asymptotes. Long division gives

\frac{x^2}{x-1} = x + 1 + \frac{1}{x-1}.

As $x \to \pm\infty$ the remainder $\frac{1}{x-1} \to 0$ , so $y = x + 1$ is a slant asymptote; there is no horizontal one. At $x = 1$ the numerator is $1 \ne 0$ , so $x = 1$ is a vertical asymptote, with $\lim_{x\to1^-} f(x) = -\infty$ and $\lim_{x\to1^+} f(x) = +\infty$ . 5. From the divided form, $f'(x) = 1 - \frac{1}{(x-1)^2} = \frac{x(x-2)}{(x-1)^2}$ . The denominator is positive, so the sign is that of $x(x-2)$ : positive on $(-\infty, 0)$ , negative on $(0,1)$ and $(1,2)$ , positive on $(2,\infty)$ . So $f$ increases on $(-\infty, 0)$ and $(2, \infty)$ , decreases on $(0,1)$ and $(1,2)$ . 6. First Derivative Test: local maximum at $x = 0$ with $f(0) = 0$ ; local minimum at $x = 2$ with $f(2) = 4$ . A local maximum lying below a local minimum is only possible because the vertical asymptote separates them. 7. $f''(x) = \frac{2}{(x-1)^3}$ , which is never zero. It is negative for $x < 1$ and positive for $x > 1$ , so the curve is concave down on $(-\infty,1)$ and concave up on $(1,\infty)$ . There is no inflection point, since $x=1$ is not in the domain. 8. The left branch rises to $(0,0)$ , turns over, and plunges to $-\infty$ along $x = 1$ , hugging the line $y = x+1$ from below far to the left. The right branch descends from $+\infty$ to the minimum $(2, 4)$ , then rises, approaching $y = x+1$ from above.□

Example 3.31 (A curve with a vertical tangent). Sketch $f(x) = 4x^{1/3} - x^{4/3}$ .

Solution.

Domain: all real $x$ , since cube roots accept negatives.
Intercepts: $f(x) = x^{1/3}(4 - x) = 0$ at $x = 0$ and $x = 4$ ; $f(0) = 0$ .
Symmetry: none.
Asymptotes: none. As $x \to +\infty$ the term $-x^{4/3}$ dominates and $f \to -\infty$ ; as $x \to -\infty$ , $x^{4/3} \to +\infty$ so $-x^{4/3} \to -\infty$ and $f \to -\infty$ again.
Differentiate:

f'(x) = \frac43 x^{-2/3} - \frac43 x^{1/3} = \frac{4(1 - x)}{3x^{2/3}}.

The denominator $3x^{2/3}$ is positive for every $x \ne 0$ , so the sign of $f'$ is the sign of $1 - x$ : $f$ increases on $(-\infty, 1)$ and decreases on $(1, \infty)$ . Critical numbers: $x = 1$ (where $f' = 0$ ) and $x = 0$ (where $f'$ is undefined). 6. At $x = 1$ , $f'$ changes from $+$ to $-$ : a local — in fact absolute — maximum, $f(1) = 3$ . At $x = 0$ there is no sign change, so no extremum; instead $f'(x) \to +\infty$ from both sides, which is a vertical tangent at the origin. 7. Second derivative:

f''(x) = -\frac89 x^{-5/3} - \frac49 x^{-2/3} = -\frac{4(x + 2)}{9x^{5/3}}.

For $x > 0$ both $x+2$ and $x^{5/3}$ are positive, so $f'' < 0$ : concave down. For $-2 < x < 0$ , $x^{5/3} < 0$ while $x + 2 > 0$ , so $f'' > 0$ : concave up. For $x < -2$ both are negative, so $f'' < 0$ : concave down. Sign changes at $x = -2$ and at $x = 0$ , so both give inflection points: $\left(-2, -6\sqrt[3]{2}\right) \approx (-2, -7.56)$ and $(0, 0)$ . 8. Coming from the far left the curve rises steeply, flexes at $x = -2$ , passes through the origin with a vertical tangent, tops out at $(1, 3)$ , crosses the axis again at $x = 4$ and falls away concave down.□

Pitfall. A slant asymptote exists only when the numerator's degree is exactly one more than the denominator's. If the degrees are equal there is a horizontal asymptote instead (the ratio of leading coefficients); if the numerator's degree exceeds the denominator's by two or more, the graph follows a parabola or higher curve and there is no linear asymptote at all. And a curve is allowed to cross a horizontal or slant asymptote — the asymptote describes behavior far out, not a barrier.

Exercises for §3.6 →

3.7Optimization problems

An optimization problem is a maximum-minimum problem dressed in a story. The calculus is the easy part; the work is turning the story into a function of a single variable, with a domain you can defend.

Method 3.32 (Steps in solving an optimization problem).

Understand the problem. Identify the quantity to be maximized or minimized, and the data given.
Draw a diagram and assign symbols to every length, area or quantity that can vary.
Write the objective — an expression for the quantity $Q$ to be optimized, in terms of those symbols.
Write the constraint — the equation the data forces on the symbols — and use it to eliminate variables until $Q$ is a function of one variable, $Q = f(x)$ .
State the domain of $f$ : the set of values of $x$ that make physical sense.
Find the absolute extremum by the Closed Interval Method if the domain is a closed interval, or by the First Derivative Test for Absolute Extrema below if it is open.
Answer the question asked — often a dimension or a price rather than the extreme value itself — and check that the answer is plausible.

Step 6 needs a tool the Extreme Value Theorem cannot supply, because an open or infinite domain has no endpoints to check.

Theorem 3.33 (First Derivative Test for Absolute Extrema). Suppose $c$ is a critical number of a continuous function $f$ defined on an interval $I$ .

If $f'(x) > 0$ for all $x < c$ in $I$ and $f'(x) < 0$ for all $x > c$ in $I$ , then $f(c)$ is the absolute maximum value of $f$ on $I$ .

If $f'(x) < 0$ for all $x < c$ in $I$ and $f'(x) > 0$ for all $x > c$ in $I$ , then $f(c)$ is the absolute minimum value of $f$ on $I$ .

Proof. Take the first case and any $x$ in $I$ with $x < c$ . The Increasing/Decreasing Test says $f$ is increasing on the part of $I$ to the left of $c$ , so $f(x) \le f(c)$ . For $x > c$ , $f$ is decreasing to the right of $c$ , so again $f(x) \le f(c)$ . Every value of $f$ on $I$ is therefore at most $f(c)$ , which is the definition of an absolute maximum. The second case is the same with the inequalities reversed.∎

The difference from the ordinary First Derivative Test is the quantifier: there the sign condition held near $c$ and the conclusion was local; here it holds on all of $I$ and the conclusion is global. When a problem produces exactly one critical number on an interval and the derivative's sign is easy to read on each side, this theorem finishes it in one line.

Intuition. Every one of these problems has the same shape. Two effects pull against each other, and the answer is the balance point. Make a fence deeper and it must get narrower; make a can taller and it needs less metal in the lids but more in the wall. The objective function adds the two effects, one of which grows with $x$ and one of which shrinks, so the sum falls then rises — and the bottom of that "V" is where the derivative is zero.

Example 3.34 (Fencing a field: maximum area). A farmer has $2400$ ft of fencing and wants to fence a rectangular field bordering a straight river, with no fence needed along the river. What dimensions give the largest area?

Solution.

Let $x$ be the width of the two sides perpendicular to the river and $y$ the length of the side parallel to it. The area is $A = xy$ .
Constraint: only three sides are fenced, so $2x + y = 2400$ , giving $y = 2400 - 2x$ .
Objective in one variable: $A(x) = x(2400 - 2x) = 2400x - 2x^{2}$ .
Domain: $x \ge 0$ and $y = 2400 - 2x \ge 0$ , so $0 \le x \le 1200$ — a closed interval.
$A'(x) = 2400 - 4x = 0$ at $x = 600$ .
Closed Interval Method: $A(0) = 0$ , $A(600) = 600 \cdot 1200 = 720{,}000$ , $A(1200) = 0$ . The maximum is $720{,}000$ ft².
Dimensions: $600$ ft by $1200$ ft. Sanity check: the fenced side parallel to the river is twice each perpendicular side, the familiar half-square answer for a three-sided rectangle.

□

Example 3.35 (A cylindrical can: minimum material). Find the dimensions of a closed cylindrical can holding $1$ L $= 1000\ \text{cm}^3$ that uses the least metal.

Solution.

Let $r$ be the radius and $h$ the height. The surface area of top, bottom and side is

S = 2\pi r^2 + 2\pi r h.

Constraint: $\pi r^{2} h = 1000$ , so $h = \dfrac{1000}{\pi r^{2}}$ .
Substituting,

S(r) = 2\pi r^2 + 2\pi r \cdot \frac{1000}{\pi r^2} = 2\pi r^2 + \frac{2000}{r}.

Domain: $r > 0$ , an open interval with no endpoints to test.
$S'(r) = 4\pi r - \dfrac{2000}{r^{2}} = \dfrac{4(\pi r^{3} - 500)}{r^{2}}$ , which is zero when $r^{3} = \dfrac{500}{\pi}$ , that is $r = \sqrt[3]{500/\pi} \approx 5.42$ cm.
For $r$ below that value $\pi r^{3} - 500 < 0$ so $S' < 0$ ; above it $S' > 0$ . By the First Derivative Test for Absolute Extrema this single critical number gives the absolute minimum on $(0,\infty)$ .
Then $h = \dfrac{1000}{\pi r^{2}} = \dfrac{1000}{\pi}\cdot\left(\dfrac{\pi}{500}\right)^{2/3}$ , and a short computation gives $h = 2r \approx 10.84$ cm. The cheapest can is exactly as tall as it is wide — a fact real cans ignore, because the lids are cut from sheet with waste and the seam costs extra.

□

Example 3.36 (Shortest distance to a curve). Find the point on the parabola $y^{2} = 2x$ closest to the point $(1, 4)$ .

Solution.

The distance from $(x,y)$ to $(1,4)$ is $d = \sqrt{(x-1)^2 + (y-4)^2}$ .
Minimizing $d$ is the same as minimizing $d^{2}$ , because $t \mapsto \sqrt{t}$ is increasing: the two have their minimum at the same point, and dropping the root avoids a messy derivative.
Use the constraint to eliminate $x$ : on the parabola $x = \tfrac{y^{2}}{2}$ , so

D(y) = \left(\frac{y^2}{2} - 1\right)^2 + (y - 4)^2.

Domain: all real $y$ .
$D'(y) = 2\left(\frac{y^{2}}{2} - 1\right)y + 2(y-4) = y^{3} - 2y + 2y - 8 = y^{3} - 8$ .
$D'(y) = 0$ when $y = 2$ . Since $D'(y) < 0$ for $y < 2$ and $D'(y) > 0$ for $y > 2$ , the First Derivative Test for Absolute Extrema gives the absolute minimum at $y = 2$ .
Then $x = \tfrac{4}{2} = 2$ , so the closest point is $(2, 2)$ , at distance $\sqrt{1 + 4} = \sqrt5$ .

□

Example 3.37 (An angle problem: the best viewing distance). A painting $3$ m high is hung on a wall with its bottom edge $1$ m above your eye level. How far from the wall should you stand so that the painting subtends the largest possible angle at your eye?

Solution.

Let $x$ be your distance from the wall and $\theta$ the angle subtended by the painting. The lines of sight to the top and bottom edges make angles $\alpha$ and $\beta$ with the horizontal, where $\tan\alpha = \frac{4}{x}$ (top edge, $1 + 3 = 4$ m up) and $\tan\beta = \frac1x$ .
Objective: $\theta = \alpha - \beta = \arctan\frac4x - \arctan\frac1x$ , for $x > 0$ .
Differentiate, using $\frac{d}{dx}\arctan u = \frac{u'}{1+u^{2}}$ :

\frac{d\theta}{dx} = \frac{-4/x^2}{1 + 16/x^2} - \frac{-1/x^2}{1 + 1/x^2} = \frac{-4}{x^2+16} + \frac{1}{x^2+1}.

Setting this to zero: $4(x^{2}+1) = x^{2} + 16$ , so $3x^{2} = 12$ and $x = 2$ (the positive root).
For $x < 2$ the expression is positive and for $x > 2$ negative, so $x = 2$ gives the absolute maximum.
Stand $2$ m from the wall. Sanity check: $2 = \sqrt{1 \cdot 4}$ , the geometric mean of the two heights — the classical answer to this problem.

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Example 3.38 (Maximizing revenue). A theatre sells $120$ seats at $\$10$ each. A survey shows that for every $\$0.50$ increase in price, $6$ fewer people attend. What ticket price maximizes revenue?

Solution.

Let $n$ be the number of $\$0.50$ increases. The price is $p = 10 + 0.5n$ dollars and attendance is $q = 120 - 6n$ .
Revenue: $R(n) = (10 + 0.5n)(120 - 6n) = 1200 - 60n + 60n - 3n^{2} = 1200 - 3n^{2}$ .
Domain: $0 \le n \le 20$ , since attendance cannot go negative.
$R'(n) = -6n$ , zero only at $n = 0$ , and $R' < 0$ for $n > 0$ .
So revenue is greatest at $n = 0$ : the price should stay at $\$10$ , for revenue $\$1200$ .

The linear terms cancelled exactly, which is the signal that the current price is already at the peak — any increase loses more customers than it gains in margin. If instead each $\$0.50$ rise cost only $4$ customers, the revenue would be $R(n) = (10+0.5n)(120-4n) = 1200 + 20n - 2n^{2}$ , maximized at $n = 5$ and a price of $\$12.50$ .□

Pitfall. Two traps recur. First, minimizing the wrong quantity: if the question asks for the dimensions, the answer is $r$ and $h$ , not the minimum surface area. Second, forgetting the domain. A critical number at $x = -3$ is irrelevant if $x$ is a length, and a problem whose domain is a closed interval is not finished until the endpoints have been compared.

Exercises for §3.7 →

3.8Related rates

In a related-rates problem, two or more quantities change with time and are tied together by an equation. One rate is known; the other is wanted. The chain rule converts the equation between the quantities into an equation between their rates, and that is the whole idea.

Step 5 before step 6 is not a stylistic preference; it is the whole method. A quantity that has been replaced by a number has derivative zero, so substituting early destroys exactly the term you were trying to compute.

Intuition. Think of the geometry as a machine with dials. The equation $x^{2} + y^{2} = 100$ for a ladder says: whatever the dials read, this relationship holds. Differentiating with respect to $t$ asks a different question — if I turn the $x$ dial at $2$ units per second, how fast must the $y$ dial turn to keep the relationship true? The chain rule is what translates "the relationship holds at every instant" into "the rates are locked together".

That also explains the surprises. The top of a ladder falls slowly while the base is near the wall and arbitrarily fast as the ladder flattens, because the same $x$ -motion demands more and more $y$ -motion to preserve $x^{2}+y^{2}=100$ .

Example 3.40 (The sliding ladder). A $10$ ft ladder rests against a vertical wall. The bottom slides away from the wall at $1$ ft/s. How fast is the top sliding down when the bottom is $6$ ft from the wall?

Solution.

Let $x$ be the distance from the wall to the bottom and $y$ the height of the top, both functions of $t$ .
Given $\frac{dx}{dt} = 1$ ft/s; wanted $\frac{dy}{dt}$ when $x = 6$ .
Pythagoras: $x^{2} + y^{2} = 100$ at every instant.
Differentiate with respect to $t$ :

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0, \qquad\text{so}\qquad \frac{dy}{dt} = -\frac{x}{y}\cdot\frac{dx}{dt}.

At the instant in question $x = 6$ , so $y = \sqrt{100-36} = 8$ , and

\frac{dy}{dt} = -\frac68 \cdot 1 = -\frac34 \text{ ft/s}.

The top slides down at $\tfrac34$ ft/s; the minus sign says $y$ is decreasing, as it must be.

Sanity check: when the bottom is at $x = 8$ (so $y = 6$ ) the formula gives $-\tfrac43$ , faster — the top accelerates downward as the ladder flattens, and as $y \to 0$ the rate is unbounded.□

Example 3.41 (Filling a conical tank). Water pours into an inverted conical tank at $2\ \text{m}^3$ /min. The tank has base radius $2$ m and height $4$ m. How fast is the water level rising when the water is $3$ m deep?

Solution.

Let $h$ be the depth of the water and $r$ the radius of its circular surface.
Given $\frac{dV}{dt} = 2\ \text{m}^3$ /min; wanted $\frac{dh}{dt}$ when $h = 3$ .
The water forms a cone: $V = \tfrac13\pi r^{2} h$ . This has two varying variables, and $\frac{dr}{dt}$ is neither given nor wanted.
Eliminate $r$ by similar triangles. The water cone is similar to the tank, so $\frac{r}{h} = \frac{2}{4}$ , that is $r = \frac h2$ . Then

V = \frac13\pi\left(\frac h2\right)^2 h = \frac{\pi h^3}{12}.

Differentiate: $\dfrac{dV}{dt} = \dfrac{\pi h^{2}}{4}\cdot\dfrac{dh}{dt}$ .
Substitute $h = 3$ and $\frac{dV}{dt} = 2$ :

2 = \frac{9\pi}{4}\cdot\frac{dh}{dt}, \qquad \frac{dh}{dt} = \frac{8}{9\pi} \approx 0.28 \text{ m/min}.

Sanity check: the surface is wider when the tank is fuller, so the same inflow raises the level more slowly — and indeed $\frac{dh}{dt}$ is proportional to $1/h^{2}$ .□

Example 3.42 (A walking man's shadow). A man $6$ ft tall walks away from a street lamp $15$ ft high at $5$ ft/s. How fast is the tip of his shadow moving, and how fast is the shadow lengthening?

Solution.

Let $x$ be the man's distance from the base of the lamp and $s$ the length of his shadow, so the tip of the shadow is at distance $x + s$ from the lamp.
Given $\frac{dx}{dt} = 5$ ft/s; wanted $\frac{d(x+s)}{dt}$ and $\frac{ds}{dt}$ .
Similar triangles — the lamp and its full triangle, the man and his — give

\frac{15}{x+s} = \frac{6}{s}.

Cross-multiplying, $15s = 6x + 6s$ , so $9s = 6x$ and $s = \tfrac23 x$ . Notice the relation holds at every instant and contains no unknown constants.
Differentiate: $\frac{ds}{dt} = \frac23 \cdot \frac{dx}{dt} = \frac23 \cdot 5 = \frac{10}{3}$ ft/s.
The tip moves at $\frac{d}{dt}(x+s) = 5 + \frac{10}{3} = \frac{25}{3} \approx 8.33$ ft/s.

The shadow lengthens at $\tfrac{10}{3}$ ft/s and its tip races ahead at $\tfrac{25}{3}$ ft/s. Both rates are constant, independent of how far the man has walked — an unusual feature of this problem and a good check that you kept the two questions apart.□

Example 3.43 (Angle of elevation of a balloon). A hot-air balloon rises vertically at $2$ m/s from a point $100$ m from an observer on the ground. How fast is the observer's angle of elevation increasing when the balloon is $100$ m high?

Solution.

Let $h$ be the balloon's height and $\theta$ the angle of elevation.
Given $\frac{dh}{dt} = 2$ m/s; wanted $\frac{d\theta}{dt}$ when $h = 100$ .
The horizontal distance is fixed at $100$ , so $\tan\theta = \dfrac{h}{100}$ .
Differentiate both sides with respect to $t$ :

\sec^2\theta \cdot \frac{d\theta}{dt} = \frac{1}{100}\cdot\frac{dh}{dt}.

When $h = 100$ the triangle is isosceles, $\theta = \tfrac\pi4$ and $\sec^{2}\theta = 2$ . Hence

2\cdot\frac{d\theta}{dt} = \frac{2}{100}, \qquad \frac{d\theta}{dt} = \frac{1}{100} = 0.01 \text{ rad/s}.

The angle grows fastest when the balloon is low and levels off as it climbs, since $\sec^{2}\theta$ grows without bound as $\theta \to \tfrac\pi2$ .□

Example 3.44 (Two vehicles approaching an intersection). Car A travels west toward an intersection at $50$ km/h and car B travels north toward the same intersection at $60$ km/h. At a certain instant A is $0.3$ km and B is $0.4$ km from the intersection. At what rate are the cars approaching each other?

Solution.

Let $x$ be A's distance from the intersection, $y$ be B's, and $z$ the distance between the cars.
Both distances are shrinking, so $\frac{dx}{dt} = -50$ and $\frac{dy}{dt} = -60$ km/h. Wanted: $\frac{dz}{dt}$ when $x = 0.3$ , $y = 0.4$ .
The roads are perpendicular, so $z^{2} = x^{2} + y^{2}$ , and at this instant $z = \sqrt{0.09 + 0.16} = 0.5$ km.
Differentiate: $2z\dfrac{dz}{dt} = 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt}$ , so $\dfrac{dz}{dt} = \dfrac{1}{z}\left(x\dfrac{dx}{dt} + y\dfrac{dy}{dt}\right)$ .
Substitute:

\frac{dz}{dt} = \frac{0.3(-50) + 0.4(-60)}{0.5} = \frac{-15 - 24}{0.5} = -78 \text{ km/h}.

The cars are approaching each other at $78$ km/h. The negative sign means $z$ is decreasing, which is exactly what "approaching" means.

Sanity check: $78$ is less than $50 + 60 = 110$ , the rate they would close at if they were on the same road — as it must be, since part of each car's motion is sideways relative to the other.□

Pitfall. Setting $x = 6$ in $x^{2} + y^{2} = 100$ before differentiating gives $36 + y^{2} = 100$ , whose $t$ -derivative is $2y\frac{dy}{dt} = 0$ and therefore $\frac{dy}{dt} = 0$ — a flatly wrong answer produced by a legal-looking step. Substitute numbers only after the differentiation. The one exception is a genuinely constant quantity, such as the ladder's length or the observer's $100$ m, which may be written as a number from the start.

Exercises for §3.8 →

3.9Motion along a line

The derivative was invented partly to describe motion, and the vocabulary of motion is the clearest place to see what the first and second derivatives mean together.

Definition 3.45 (Velocity, speed and acceleration). If a particle moves along a straight line with position $s(t)$ at time $t$ , its velocity, speed and acceleration are

v(t) = s'(t), \qquad \text{speed} = |v(t)|, \qquad a(t) = v'(t) = s''(t).

Velocity is signed: $v(t) > 0$ means the particle is moving in the positive direction, $v(t) < 0$ in the negative direction, and $v(t) = 0$ means it is momentarily at rest. Speed discards the sign and reports only how fast. Acceleration is the rate of change of velocity, which is not the same as the rate of change of speed — a distinction that settles the most-asked question about motion problems.

Proposition 3.46 (When a particle speeds up). A particle speeds up when $v(t)$ and $a(t)$ have the same sign, and slows down when they have opposite signs.

Proof. Speed is $|v(t)|$ . Where $v > 0$ , speed equals $v$ , so its rate of change is $a$ : the speed increases exactly when $a > 0$ , that is when $a$ has the same sign as $v$ . Where $v < 0$ , speed equals $-v$ , so its rate of change is $-a$ , which is positive exactly when $a < 0$ — again the same sign as $v$ .∎

A particle changes direction at a time when $v$ changes sign; if $v$ touches zero without changing sign, the particle merely pauses. Over an interval $[a,b]$ , the net change of position is the displacement $s(b) - s(a)$ , while the total distance travelled adds up the unsigned motion and is found by splitting $[a,b]$ at the times when $v = 0$ and summing $|s(t_{i+1}) - s(t_i)|$ over the pieces. (The integrals chapter writes these as $\int_a^b v\,dt$ and $\int_a^b |v|\,dt$ .)

Intuition. Sit in a car facing forward. Velocity is the speedometer plus a sign for whether you are in drive or reverse. Acceleration is the pedal: positive means pressing the accelerator in the forward direction, negative means pressing the brake or reversing.

Reversing down a driveway with your foot pressing harder, velocity is negative and acceleration is negative too — same sign, and you are indeed speeding up, even though both numbers are negative. That is why "negative acceleration" does not mean "slowing down".

Example 3.47 (A full analysis of a motion). A particle moves with position $s(t) = t^{3} - 6t^{2} + 9t$ metres for $t \ge 0$ . Find when it is at rest, when it moves forward, when it is speeding up, and the total distance travelled in the first $5$ s.

Solution.

$v(t) = 3t^{2} - 12t + 9 = 3(t-1)(t-3)$ and $a(t) = 6t - 12 = 6(t - 2)$ .
At rest when $v = 0$ : $t = 1$ and $t = 3$ s.
Sign of $v$ : positive on $[0,1)$ , negative on $(1,3)$ , positive on $(3,\infty)$ . So the particle moves forward on $[0,1)$ and after $t = 3$ , backward between.
Sign of $a$ : negative for $t < 2$ , positive for $t > 2$ .
Same signs — speeding up — on $(1,2)$ (both negative) and on $(3,\infty)$ (both positive). Opposite signs — slowing down — on $(0,1)$ and $(2,3)$ .
Total distance: $s(0) = 0$ , $s(1) = 1 - 6 + 9 = 4$ , $s(3) = 27 - 54 + 27 = 0$ , $s(5) = 125 - 150 + 45 = 20$ . The legs measure $|4 - 0| = 4$ , $|0 - 4| = 4$ and $|20 - 0| = 20$ , so the total distance is $28$ m, while the displacement is only $20$ m.

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Example 3.48 (Projectile motion). A ball is thrown upward from ground level with initial velocity $64$ ft/s, so that $h(t) = -16t^{2} + 64t$ . Find its maximum height, its velocity on impact, and its speed at $t = 3$ s.

Solution.

$v(t) = h'(t) = -32t + 64$ , and $a(t) = -32$ ft/s² — constant, the acceleration due to gravity.
Maximum height where $v = 0$ : $t = 2$ s, and $h(2) = -64 + 128 = 64$ ft. The Second Derivative Test confirms a maximum since $a = -32 < 0$ .
Impact when $h = 0$ : $-16t(t - 4) = 0$ , so $t = 4$ s. Then $v(4) = -128 + 64 = -64$ ft/s: it returns with the same speed it left, directed downward.
At $t = 3$ : $v(3) = -96 + 64 = -32$ ft/s, so the speed is $32$ ft/s. Since $v < 0$ and $a < 0$ agree in sign, the ball is speeding up as it falls.

□

Exercises for §3.9 →

3.10Linear approximation and differentials

The tangent line was built to describe a curve at one point. Because it stays close to the curve for a while on either side, it can also stand in for the curve — and a line is something you can evaluate in your head.

Definition 3.49 (Linearization). If $f$ is differentiable at $a$ , the linearization of $f$ at $a$ is the function

L(x) = f(a) + f'(a)(x - a),

and the approximation $f(x) \approx L(x)$ for $x$ near $a$ is called the linear approximation, or tangent line approximation, of $f$ at $a$ .

Why it works is the definition of the derivative read backwards. Since

f'(a) = \lim_{x\to a}\frac{f(x)-f(a)}{x-a},

the difference between the true value and the linear one,

f(x) - L(x) = \left(\frac{f(x)-f(a)}{x-a} - f'(a)\right)(x-a),

is a product of two factors that both tend to $0$ . So the error vanishes faster than $x - a$ itself — for twice-differentiable $f$ it is proportional to $(x-a)^2$ , a fact the Taylor series makes exact. Halve your distance from $a$ and the error drops by about a factor of four.

The base point $a$ must be chosen so that $f(a)$ and $f'(a)$ are known exactly and $a$ is close to the value you want. To estimate $\sqrt{4.02}$ take $a = 4$ ; to estimate $\sqrt{101}$ take $a = 100$ , not $a = 4$ .

Notation. Leibniz's differentials repackage the same statement. If $y = f(x)$ , let $dx$ be an independent variable — any real number — and define the differential $dy$ by

dy = f'(x)\,dx.

Then $dy$ is the change in height along the tangent line when $x$ changes by $dx$ , while $\Delta y = f(x + dx) - f(x)$ is the change along the curve. The linear approximation says $\Delta y \approx dy$ for small $dx$ . This is also why $\frac{dy}{dx}$ can be treated as a quotient in the substitution rule later.

Intuition. Zoom in on any smooth curve far enough and it becomes indistinguishable from a straight line — that is what differentiable means. Linearization says: stop zooming, keep the line, and use it for anything close to where you zoomed.

Concretely, $\sqrt{4} = 2$ is exact and the slope there is $\frac{1}{2\sqrt4} = 0.25$ . So $\sqrt{4.02} \approx 2 + 0.25(0.02) = 2.005$ . The true value is $2.004994\ldots$ — an error of six millionths, from arithmetic you can do while walking.

Example 3.50 (Estimating a cube root). Use a linear approximation to estimate $\sqrt[3]{8.1}$ , and estimate the error.

Solution.

Take $f(x) = x^{1/3}$ and $a = 8$ , since $f(8) = 2$ exactly.
$f'(x) = \tfrac13 x^{-2/3}$ , so $f'(8) = \tfrac13 \cdot \tfrac14 = \tfrac1{12}$ .
$L(x) = 2 + \tfrac{1}{12}(x - 8)$ .
$\sqrt[3]{8.1} \approx L(8.1) = 2 + \tfrac{0.1}{12} = 2.008\overline{3}$ .
The true value is $2.00829885\ldots$ , so the error is about $4\times10^{-5}$ — and the curve is concave down here ( $f'' < 0$ ), which is why the tangent line estimate comes out slightly too large.

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Example 3.51 (Error propagation with differentials). The radius of a sphere is measured as $21$ cm with a possible error of at most $0.05$ cm. Estimate the resulting error in the computed volume, and the relative error.

Solution.

$V = \tfrac43\pi r^{3}$ , so $dV = 4\pi r^{2}\,dr$ .
With $r = 21$ and $dr = 0.05$ :

dV = 4\pi (21)^2(0.05) = 4\pi(441)(0.05) = 88.2\pi \approx 277\ \text{cm}^3.

The computed volume is $V = \tfrac43\pi(21)^3 = 12348\pi \approx 38792\ \text{cm}^3$ , so the relative error is

\frac{dV}{V} = \frac{4\pi r^2\,dr}{\tfrac43\pi r^3} = 3\,\frac{dr}{r} = 3\cdot\frac{0.05}{21} \approx 0.0071,

about $0.7\%$ .

An absolute error of $277\ \text{cm}^3$ sounds alarming and is in fact tiny. Note the general rule the algebra exposed: a relative error in a radius is tripled in a volume, because volume is the third power of length.□

Pitfall. A linear approximation is only trustworthy near its base point, and "near" is measured against the curvature. Estimating $\sqrt{100}$ from the linearization of $\sqrt x$ at $a = 4$ gives $2 + 0.25(96) = 26$ , against the true value $10$ . The error term grows like $(x-a)^2$ , so moving $24$ times further from the base point makes the error roughly $576$ times worse.

Exercises for §3.10 →

3.11Newton's method

Most equations cannot be solved in closed form. There is no formula for the root of $x^{3} + x - 1 = 0$ that a calculator could not have found faster numerically, and no formula at all for $\cos x = x$ . Newton's method turns the linear approximation into a root-finding machine.

Suppose $r$ is a root of $f(x) = 0$ and $x_n$ is a guess near it. Replace $f$ by its linearization at $x_n$ ,

L(x) = f(x_n) + f'(x_n)(x - x_n),

and solve the easy equation $L(x) = 0$ instead. That gives $x = x_n - \dfrac{f(x_n)}{f'(x_n)}$ , and calling the result the next guess produces the method.

Method 3.52 (Newton's method). To approximate a root of $f(x) = 0$ , choose an initial guess $x_0$ and iterate

x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)},

provided $f'(x_n) \ne 0$ . Stop when successive iterates agree to the accuracy required, that is when $|x_{n+1} - x_n|$ is smaller than the tolerance, and confirm by checking that $|f(x_{n+1})|$ is small.

Geometrically, each step draws the tangent line to the curve at the current guess and slides down it to the $x$ -axis. Where the curve is nearly straight, the tangent nearly is the curve, and the landing point is nearly the root.

Proposition 3.53 (Quadratic convergence). If $f''$ is continuous, $f'(r) \ne 0$ at the root $r$ , and $x_0$ is chosen close enough to $r$ , then the errors $e_n = x_n - r$ satisfy

|e_{n+1}| \approx \frac{|f''(r)|}{2|f'(r)|}\,|e_n|^2.

The full proof is a Taylor-expansion argument that belongs with the error analysis of Taylor series; the idea is that expanding $f$ around $r$ to second order and substituting the iteration formula makes the first-order terms cancel exactly, leaving a term in $e_n^{2}$ . The practical meaning is the memorable part: squaring a small error roughly doubles the number of correct decimal places at each step. Three correct digits become six, then twelve.

Intuition. You are standing on a hillside above a valley whose lowest point touches sea level, and you want to find where the ground meets the sea. Look at the slope right where you stand, assume the hill keeps that slope, and walk to where that straight line would hit sea level. You overshoot or undershoot, but you are much closer. Look at the new slope and repeat. Each look corrects most of the remaining error, which is why the method converges so fast — and also why it fails when the slope where you stand points somewhere misleading.

Example 3.54 (Computing a cube root). Use Newton's method to approximate $\sqrt[3]{2}$ to six decimal places, starting from $x_0 = 1$ .

Solution.

$\sqrt[3]{2}$ is the root of $f(x) = x^{3} - 2$ , with $f'(x) = 3x^{2}$ .
The iteration is

x_{n+1} = x_n - \frac{x_n^3 - 2}{3x_n^2} = \frac{2x_n^3 + 2}{3x_n^2}.

$x_0 = 1$ gives $x_1 = \dfrac{2 + 2}{3} = 1.333333$ .
$x_2 = \dfrac{2(1.333333)^3 + 2}{3(1.333333)^2} = \dfrac{6.740741}{5.333333} = 1.263889$ .
$x_3 = 1.259933$ , $x_4 = 1.259921$ .
$x_3$ and $x_4$ agree to four decimals and $x_4$ is correct to eight: $\sqrt[3]{2} = 1.25992105\ldots$

Count the correct digits: $1$ , then $2$ , then $5$ , then $8$ . That doubling is quadratic convergence.□

Example 3.55 (A failure: divergence). Apply Newton's method to $f(x) = x^{1/3}$ , whose only root is $x = 0$ , starting from $x_0 = 1$ .

Solution.

$f'(x) = \tfrac13 x^{-2/3}$ , so

x_{n+1} = x_n - \frac{x_n^{1/3}}{\tfrac13 x_n^{-2/3}} = x_n - 3x_n = -2x_n.

From $x_0 = 1$ the iterates are $-2, 4, -8, 16, \ldots$ : the guesses alternate in sign and double in size, running away from the root that sits at $0$ .
The cause is visible in the graph. At the root the tangent is vertical — $f'(0)$ does not exist — so the hypothesis $f'(r) \ne 0$ of the convergence result fails, and each tangent line crosses the axis further out than the point it was drawn at.

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Pitfall. Newton's method can fail in three ways beyond the one just seen. If $f'(x_n) = 0$ the tangent is horizontal and the next iterate does not exist — the method divides by zero. If the initial guess is poor the iterates may converge to a different root than the one you wanted, or wander off. And the iteration can cycle: for $f(x) = x^{3} - 2x + 2$ starting at $x_0 = 0$ one computes $x_1 = 1$ , then $x_2 = 0$ , and the sequence oscillates between $0$ and $1$ forever. A quick sketch, or a bisection step or two to get close first, prevents most of this.

Exercises for §3.11 →

3.12Antiderivatives

Every application so far has started from $f$ and computed $f'$ . A physicist usually has the opposite problem: acceleration is known, and position is wanted. That reverses the arrow.

Definition 3.56 (Antiderivative). A function $F$ is called an antiderivative of $f$ on an interval $I$ if $F'(x) = f(x)$ for all $x$ in $I$ .

Theorem 3.57 (The general antiderivative). If $F$ is an antiderivative of $f$ on an interval $I$ , then the most general antiderivative of $f$ on $I$ is $F(x) + C$ , where $C$ is an arbitrary constant.

Proof. Each $F + C$ has derivative $F' = f$ , so all of them are antiderivatives. Conversely, if $G$ is any antiderivative then $G' = f = F'$ on $I$ , so by the corollary on equal derivatives, $G - F$ is constant on $I$ : $G = F + C$ for some constant $C$ .∎

Reversing each differentiation rule gives the basic table — $x^n$ has antiderivative $\frac{x^{n+1}}{n+1}$ for $n \ne -1$ , $\frac1x$ has $\ln|x|$ , $\cos x$ has $\sin x$ , $\sin x$ has $-\cos x$ , $\sec^{2} x$ has $\tan x$ , $e^x$ has $e^x$ , and $\frac{1}{1+x^{2}}$ has $\arctan x$ — each up to an additive constant. Antidifferentiation is also linear: an antiderivative of $cf + g$ is $c$ times one of $f$ plus one of $g$ . The integrals chapter takes this up properly; here it is the tool that turns a rate back into a quantity.

Example 3.58 (Recovering position from acceleration). A particle moves with acceleration $a(t) = 6t + 4$ , initial velocity $v(0) = -6$ and initial position $s(0) = 9$ . Find $s(t)$ .

Solution.

$v$ is an antiderivative of $a$ : $v(t) = 3t^{2} + 4t + C$ .
The initial condition $v(0) = -6$ gives $C = -6$ , so $v(t) = 3t^{2} + 4t - 6$ .
$s$ is an antiderivative of $v$ : $s(t) = t^{3} + 2t^{2} - 6t + D$ .
The condition $s(0) = 9$ gives $D = 9$ , so $s(t) = t^{3} + 2t^{2} - 6t + 9$ .
Check by differentiating twice: $s' = 3t^{2} + 4t - 6 = v$ and $s'' = 6t + 4 = a$ .

Two antidifferentiations needed two initial conditions. That pattern — one arbitrary constant per integration, one condition to pin it down — is the whole theory of initial value problems in miniature.□

Exercises for §3.12 →

3.13Rates of change in the sciences

The derivative is the general-purpose language for "how fast". Only the names of the variables change from field to field, and recognising that saves relearning the calculus in each one.

In physics, if $s(t)$ is position then $v = s'$ is velocity and $a = v' = s''$ is acceleration, and Newton's second law reads $F = ma = ms''(t)$ . If $Q(t)$ is charge through a wire, the current is $I = Q'(t)$ ; if $m(x)$ is the mass of a rod from one end to the point $x$ , the linear density is $\rho = m'(x)$ .

In chemistry, the rate of a reaction is $-\frac{d[A]}{dt}$ , the rate at which the concentration of reactant $A$ falls, and a rate law such as $-\frac{d[A]}{dt} = k[A]^n$ relates it to concentration. The compressibility of a gas is $-\frac{1}{V}\frac{dV}{dP}$ , the relative change in volume per unit change in pressure.

In biology, if $P(t)$ is a population then $P'(t)$ is the growth rate and $\frac{P'(t)}{P(t)}$ is the per-capita or relative growth rate. Constant relative growth gives $P' = kP$ and exponential growth; the logistic model $P' = kP\left(1 - \frac{P}{K}\right)$ bends that to a carrying capacity $K$ , and $P = K/2$ is exactly the inflection point of the population curve.

In economics, the marginal cost is $C'(x)$ , the cost of one more unit; marginal revenue is $R'(x)$ ; and profit $P = R - C$ is maximized where $P' = 0$ , that is where $R'(x) = C'(x)$ — marginal revenue equals marginal cost, with the Second Derivative Test requiring $R'' < C''$ to confirm it is a maximum and not a minimum.

Intuition. Every one of these is the same sentence with the nouns swapped: "the derivative of a stock is its flow". Position and velocity, charge and current, mass and density, population and births per year, total cost and cost of the next unit. When you meet a new field, the question to ask is only which quantity is the stock and which the flow.

Example 3.59 (Marginal cost and average cost). A firm's cost function is $C(x) = 2000 + 3x + 0.01x^{2}$ dollars for $x$ units. Find the marginal cost at $x = 100$ , the actual cost of the $101$ st unit, and the production level that minimizes average cost.

Solution.

Marginal cost is $C'(x) = 3 + 0.02x$ , so $C'(100) = 5$ dollars per unit.
The actual cost of the $101$ st unit is $C(101) - C(100) = (2000 + 303 + 102.01) - (2000 + 300 + 100) = 5.01$ dollars. The marginal cost approximates it, and the small discrepancy is exactly the linear approximation error.
Average cost is $c(x) = \dfrac{C(x)}{x} = \dfrac{2000}{x} + 3 + 0.01x$ for $x > 0$ .
$c'(x) = -\dfrac{2000}{x^{2}} + 0.01 = 0$ gives $x^{2} = 200000$ , so $x = \sqrt{200000} \approx 447$ units.
$c'$ is negative below and positive above that value, so by the First Derivative Test for Absolute Extrema it is the absolute minimum. There $c \approx 4.47 + 3 + 4.47 = 11.94$ dollars per unit.

Notice that $C'(447) = 3 + 8.94 = 11.94$ as well: average cost is minimized exactly where marginal cost equals average cost, which is true for every differentiable cost function.□

Example 3.60 (Relative growth rate). A bacterial culture grows so that $P(t) = 500e^{0.03t}$ thousand cells after $t$ hours. Find the growth rate at $t = 10$ and the relative growth rate at any time.

Solution.

$P'(t) = 500(0.03)e^{0.03t} = 15e^{0.03t}$ thousand cells per hour.
At $t = 10$ : $P'(10) = 15e^{0.3} \approx 20.2$ thousand cells per hour.
Relative growth rate: $\dfrac{P'(t)}{P(t)} = \dfrac{15e^{0.03t}}{500e^{0.03t}} = 0.03$ , a constant $3\%$ per hour.

The absolute growth rate climbs forever while the relative rate never changes — the defining signature of exponential growth, and the reason biologists quote the percentage rather than the raw number.□

Summary. A continuous $f$ on a closed interval $[a,b]$ attains an absolute maximum and minimum (Extreme Value Theorem), and Fermat's Theorem says an interior extremum can only sit at a critical number, where $f'(c)=0$ or $f'(c)$ fails to exist — hence the Closed Interval Method: compare $f$ at the critical numbers in $(a,b)$ and at the two endpoints. The Mean Value Theorem, $f(b)-f(a)=f'(c)(b-a)$ for $f$ continuous on $[a,b]$ and differentiable on $(a,b)$ , proves everything after it: $f'>0$ means increasing and $f'<0$ decreasing; the First Derivative Test reads a local maximum off a sign change of $f'$ from $+$ to $-$ (and an absolute one when the sign change governs the whole interval); $f''>0$ means concave upward, $f''<0$ concave downward, with an inflection point where $f''$ actually changes sign; and if $f''$ is continuous near $c$ with $f'(c)=0$ , then $f''(c)>0$ gives a local minimum and $f''(c)<0$ a local maximum, with $f''(c)=0$ silent. Locally, $f(x)\approx L(x)=f(a)+f'(a)(x-a)$ with error of order $(x-a)^2$ , which iterated as $x_{n+1}=x_n-f(x_n)/f'(x_n)$ is Newton's method, converging quadratically when $f'(r)\ne 0$ . Finally $F'=f$ on an interval makes $F+C$ the general antiderivative.

Exercises for §3.13 →

Common Mistakes

Treating every critical number as an extremum. Fermat's Theorem runs one way only: $f'(c) = 0$ at a local extremum, but $x^{3}$ has $f'(0) = 0$ and no extremum there. Always apply the First or Second Derivative Test.
Forgetting the critical numbers where $f'$ fails to exist. A cusp or a corner can carry a maximum, and $f'(x) = 0$ will never find it.
Forgetting the endpoints in a closed-interval problem, or checking endpoints on an interval that has none. Closed interval: compare critical values and endpoint values. Open interval: use the First Derivative Test for Absolute Extrema instead.
Concluding " $f''(c) = 0$ , therefore inflection point". The concavity must actually change: $x^{4}$ has $f''(0) = 0$ and is concave up on both sides.
Using the Second Derivative Test when $f''(c) = 0$ . It is inconclusive, not negative — fall back on the First Derivative Test.
Saying " $f' = 0$ everywhere, so $f$ is constant" on a domain that is not an interval. The corollary needs an interval; $x/|x|$ is the counterexample.
Plugging numbers in before differentiating in a related-rates problem. A quantity replaced by a constant has derivative zero, which silently deletes the term you wanted.
Confusing velocity with speed, and reading "negative acceleration" as "slowing down". A particle speeds up whenever $v$ and $a$ share a sign.
In optimization, answering with the extreme value when the question asked for the dimensions, or forgetting to check that the variable's domain excludes negative lengths.
Trusting a linear approximation far from its base point. The error grows like the square of the distance, so doubling the distance quadruples it.
Applying Newton's method without a sketch. A guess near a horizontal tangent, or near a root with a vertical tangent, can diverge or cycle no matter how many iterations you run.
Writing an antiderivative without $+C$ , then being surprised that an initial condition cannot be satisfied.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

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