Partial Derivatives

(a) Two restrictions, then intersect. The square root needs $y-x\ge 0$, that is $y\ge x$; the denominator needs $x^2+y^2\ne 1$. So

$$D=\{(x,y):y\ge x,x^2+y^2\ne 1\}.$$

Draw the line $y=x$ solid, since points on it satisfy $y-x=0\ge 0$ and belong to $D$, and shade the closed half-plane above and to the left of it (test point $(0,1)$: $1-0\ge 0$). Then remove the unit circle. Only the part of the circle inside the shaded region matters: the arc running counter-clockwise from $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ through $(0,1)$ and $(-1,0)$ to $\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$. Its two endpoints are where $y=x$ meets $x^2+y^2=1$, so $2x^2=1$; they lie on the circle and are therefore excluded, even though they sit on the included line. Draw the arc dashed and its endpoints as open circles. The rest of the circle lies below the line and was never in the domain.

(b) Solve $g=k$ for $y$. $g(x,y)=k$ means $y=k\left(x^2+1\right)$. For $k=0$ this is the $x$-axis. For $k=1$ it is the parabola $y=x^2+1$, vertex $(0,1)$, opening upward; for $k=2$ it is $y=2x^2+2$, vertex $(0,2)$, opening upward and narrower; for $k=-1$ it is $y=-x^2-1$, vertex $(0,-1)$, opening downward. No two of them meet, as level curves for different values never can.

Read the spacing. Along the $y$-axis the four curves cross at $y=-1,0,1,2$, one unit apart. Along the line $x=1$ they cross at $y=-2,0,2,4$, two units apart, and along $x=2$ at $y=-5,0,5,10$. The contours are packed tightest on the $y$-axis and spread out as $|x|$ grows, so the surface is steepest above the $y$-axis and flattens away from it.

From the origin, which lies on the curve $k=0$, the nearest point of the curve $k=1$ is its vertex $(0,1)$, at distance $1$: any other point $\left(x,x^2+1\right)$ of that curve is at distance $\sqrt{x^2+{\left(x^2+1\right)}^2}>1$. So the shortest walk that gains one unit of $g$ is straight up the positive $y$-axis, and that is the direction of fastest increase. (The gradient $\nabla g(0,0)=⟨0,1⟩$ agrees.)

(c) Solve for $z$. $F=k$ means $z=\sqrt{x^2+y^2}+k$. For $k=0$ this is $z=\sqrt{x^2+y^2}$: the upper half of the cone $z^2=x^2+y^2$, vertex at the origin, opening upward, with its sides at $45^{\circ }$ to the axis (in the plane $y=0$ it is $z=|x|$); its horizontal cross-section at height $z=c>0$ is a circle of radius $c$. For $k=1$ the same half-cone is shifted up one unit, vertex $(0,0,1)$; for $k=-1$ it is shifted down, vertex $(0,0,-1)$. Every level surface is a vertical translate of the single half-cone $z=\sqrt{x^2+y^2}$, and since $F$ is defined on all of ${\mathbb{R}}^3$ with range all of $\mathbb{R}$, every real $k$ produces one.

(a) Bound it in polar coordinates. Substitution gives $\frac{0}{0}$. With $x=r\cos \theta$, $y=r\sin \theta$,

$$\frac{x^3-y^3}{x^2+y^2}=\frac{r^3\left({\cos }^3\theta -{\sin }^3\theta \right)}{r^2}=r\left({\cos }^3\theta -{\sin }^3\theta \right).$$

Since $|{\cos }^3\theta -{\sin }^3\theta |\le |\cos \theta {|}^3+|\sin \theta {|}^3\le 2$ for every $\theta$,

$$\left|\frac{x^3-y^3}{x^2+y^2}-0\right|\le 2r,$$

which tends to $0$ as $r\to 0^{+}$ regardless of $\theta$. The limit is $0$; in $\epsilon$–$\delta$ terms, $\delta =\epsilon /2$ works. Without polar coordinates: $\frac{x^2}{x^2+y^2}\le 1$ gives $\frac{|x{|}^3}{x^2+y^2}\le |x|$, likewise for $y$, so $|f(x,y)|\le |x|+|y|\to 0$.

(b) Every line agrees; a cubic curve does not. Along $y=mx$,

$$f(x,mx)=\frac{x\cdot m^3{x}^3}{x^2+m^6{x}^6}=\frac{m^3{x}^2}{1+m^6{x}^4}⟶0,$$

and along the $y$-axis $f(0,y)=0$. Every straight line reports $0$. But along the curve $x=y^3$,

$$f\left(y^3,y\right)=\frac{y^3\cdot y^3}{y^6+y^6}=\frac{1}{2}(y\ne 0).$$

Two paths into the origin, limits $0$ and $\frac{1}{2}$: by the two-path test the limit does not exist. Polar coordinates would not have rescued this one — after substitution the expression still depends on $\theta$ in a way no bound in $r$ alone controls, which is itself the warning sign.

(c) Factor and cancel. At $(1,2)$ the numerator is $2-2-2+2=0$ and the denominator is $0$: the form is $\frac{0}{0}$. Group the numerator:

$$x^{2}y-2x^2-y+2=x^2(y-2)-(y-2)=\left(x^2-1\right)(y-2)=(x-1)(x+1)(y-2).$$

The function is undefined on the whole line $x=1$, but a limit only looks at points of the domain, and at those points $x-1\ne 0$ may be cancelled:

$$\frac{x^{2}y-2x^2-y+2}{x-1}=(x+1)(y-2)(x\ne 1).$$

The right-hand side is a polynomial, hence continuous, so its limit is its value: $(1+1)(2-2)=0$.

(a) Product rule with the chain rule inside. Holding $y$ constant, $x^2\cos (xy)$ is a product of $x^2$ and $\cos (xy)$, and $\frac{\partial }{\partial x}(xy)=y$:

$$f_x=2x\cos (xy)+x^2\cdot (-\sin (xy))\cdot y=2x\cos (xy)-x^{2}y\sin (xy).$$

Holding $x$ constant, $x^2$ is a constant factor and $\frac{\partial }{\partial y}(xy)=x$:

$$f_y=x^2\cdot (-\sin (xy))\cdot x=-x^3\sin (xy).$$

(b) Differentiate twice in each variable. With $y$ held constant, $u_x=\frac{2x}{x^2+y^2}$, and by the quotient rule

$$u_{xx}=\frac{2\left(x^2+y^2\right)-2x\cdot 2x}{{\left(x^2+y^2\right)}^2}=\frac{2y^2-2x^2}{{\left(x^2+y^2\right)}^2}.$$

The formula for $u$ is unchanged when $x$ and $y$ are swapped, so swapping them in $u_{xx}$ gives $u_{yy}=\frac{2x^2-2y^2}{{\left(x^2+y^2\right)}^2}$ (a direct computation from $u_y=\frac{2y}{x^2+y^2}$ confirms it). Adding,

$$u_{xx}+u_{yy}=\frac{2y^2-2x^2+2x^2-2y^2}{{\left(x^2+y^2\right)}^2}=0.$$

Every step needed $x^2+y^2\ne 0$, which is exactly the punctured plane on which $u$ is defined. So $u$ is harmonic there.

(c) Choose the order with Clairaut's Theorem. Every partial derivative of $g$, of every order, is built from polynomials, $\sin y$, $e^{2x}$ and the quotient $\frac{x^5}{1+e^x}$ whose denominator never vanishes, so they are all continuous on ${\mathbb{R}}^2$. Clairaut's Theorem, applied twice, gives $g_{xyy}=g_{yxy}=g_{yyx}$: only the number of differentiations in each variable matters. Differentiate in $y$ first, because the term $\frac{x^5}{1+e^x}$ contains no $y$ and vanishes at the first stroke, whereas differentiating it in $x$ first would cost a quotient rule for nothing:

$$g_y=x\cos y+3y^2{e}^{2x},g_{yy}=-x\sin y+6y{e}^{2x},g_{yyx}=-\sin y+12y{e}^{2x}.$$

Hence $g_{xyy}=-\sin y+12y{e}^{2x}$. Check by the other order on the two surviving terms: $g_x=\sin y+2y^3{e}^{2x}+(\text{a function of }x\text{ alone})$, $g_{xy}=\cos y+6y^2{e}^{2x}$, $g_{xyy}=-\sin y+12y{e}^{2x}$, the same.

(a) Two partials, one formula. $f_x=2x{e}^y$ and $f_y=x^2{e}^y$ are continuous on all of ${\mathbb{R}}^2$, which is the hypothesis the tangent plane theorem needs. At $(2,0)$: $f=4$, $f_x=4$, $f_y=4$. The tangent plane is

$$z-4=4(x-2)+4(y-0),\text{that is,}z=4x+4y-4.$$

Check: at $(2,0)$ the plane gives $8+0-4=4$, the height of the point.

(b) Evaluate the plane, not the function. The linearization is $L(x,y)=4x+4y-4$, so

$$f(1.9,0.05)\approx L(1.9,0.05)=7.6+0.2-4=\mathrm{3.8.}$$

(For reference, the true value is $3.61e^{0.05}\approx 3.795$; the step from the base point is small and the estimate is good to two decimals.)

(c) Differential, worst case. $dz=f_{x}dx+f_{y}dy=2x{e}^{y}dx+x^2{e}^{y}dy$, which at $(2,0)$ is $dz=4dx+4dy$. With $|dx|\le 0.01$ and $|dy|\le 0.02$, the worst case is both errors with the same sign:

$$|dz|\le 4(0.01)+4(0.02)=0.04+0.08=\mathrm{0.12.}$$

The maximum relative error is $\frac{0.12}{4}=0.03$, that is $3\%$. The same number comes from dividing through: $\frac{dz}{z}=\frac{2dx}{x}+dy=2(0.005)+0.02=0.03$. Note that an absolute error in $y$ becomes a relative error in $f$, because $y$ sits in the exponent.

(a) Cancel first, then estimate. For $(x,y)\ne (0,0)$ the denominator is nonzero and

$$\frac{x^4-y^4}{x^2+y^2}=\frac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{x^2+y^2}=x^2-y^2,$$

and $|x^2-y^2|\le x^2+y^2=r^2$.

Proof. Let $\epsilon >0$ and set $\delta =\sqrt{\epsilon }$. If $0<r<\delta$, then

$$\left|\frac{x^4-y^4}{x^2+y^2}-0\right|=|x^2-y^2|\le r^2<{\delta }^2=\epsilon .$$

Since $\epsilon$ was arbitrary, the limit is $0$. $■$

(b) Only $c=0$. Away from the origin $f$ is a rational function whose denominator does not vanish, and rational functions are continuous wherever they are defined (limits of sums, products and quotients are the sums, products and quotients of the limits, with $\lim x=a$ and $\lim y=b$ from the definition). At the origin, continuity means ${\lim }_{(x,y)\to (0,0)}f(x,y)=f(0,0)$; by (a) the left side is $0$, so continuity holds exactly when $c=0$. A second way to see it: with $c=0$ the function is $f(x,y)=x^2-y^2$ at every point, the origin included, and that is a polynomial.

(c) Proof. Assume, as the statement implicitly does, that $f$ is defined at the points of the line near $(a,b)$. Let $\epsilon >0$. By hypothesis there is $\delta >0$ such that $|f(x,y)-L|<\epsilon$ whenever $0<r<\delta$. A point of the line has the form $(x,b+m(x-a))$, and its distance from $(a,b)$ is

$$r=\sqrt{(x-a{)}^2+m^2(x-a{)}^2}=|x-a|\sqrt{1+m^2}.$$

Put ${\delta }^{\prime }=\frac{\delta }{\sqrt{1+m^2}}>0$. If $0<|x-a|<{\delta }^{\prime }$, then $0<r=|x-a|\sqrt{1+m^2}<\delta$, so $\left|f\right(x,b+m(x-a))-L|<\epsilon$. That is precisely the $\epsilon$–$\delta$ statement of ${\lim }_{x\to a}f(x,b+m(x-a))=L$, with ${\delta }^{\prime }$ as the tolerance. $■$

The two-path test is the contrapositive: if the limit existed, every line (and, by the same argument with the distance along any curve into $(a,b)$, every path) would have to report the same value $L$, so two paths reporting different values rule the limit out. Note the direction of the implication — the result never says that lines agreeing makes the limit exist.

(a) Step 2 is the first error. Step 1 is correct: $f(h,0)=\frac{0}{h^4}=0$ and $f(0,k)=\frac{0}{k^2}=0$, so both difference quotients are identically $0$. Step 2 appeals to the theorem continuous partials imply differentiability: if $f_x$ and $f_y$ exist near $(a,b)$ and are continuous at $(a,b)$, then $f$ is differentiable at $(a,b)$. The hypothesis is continuity of the partial derivatives, not their existence, and Step 1 established only two values. Existence of $f_x(a,b)$ and $f_y(a,b)$ says nothing about differentiability, because the partials look along two lines only while differentiability is a statement about every direction at once.

The hypothesis in fact fails here. For $(x,y)\ne (0,0)$ the quotient rule gives

$$f_x=\frac{2xy\left(x^4+y^2\right)-x^{2}y\cdot 4x^3}{{\left(x^4+y^2\right)}^2}=\frac{2x{y}^3-2x^{5}y}{{\left(x^4+y^2\right)}^2},$$

and along $y=2x^2$ this is $\frac{16x^7-4x^7}{25x^8}=\frac{12}{25x}$, which is unbounded as $x\to 0$. So $f_x$ is not continuous at the origin, and the theorem is silent. (Part (b) gives a shorter reason: were the partials continuous, $f$ would be differentiable, hence continuous, and it is not.)

(b) Two paths. Along the $x$-axis, $f(x,0)=0\to 0$. Along the parabola $y=x^2$,

$$f\left(x,x^2\right)=\frac{x^2\cdot x^2}{x^4+x^4}=\frac{1}{2}(x\ne 0),$$

so $f\to \frac{1}{2}$ on that path. By the two-path test ${\lim }_{(x,y)\to (0,0)}f(x,y)$ does not exist, and continuity at the origin — which requires that limit to exist and equal $f(0,0)=0$ — fails. The corollary differentiability implies continuity, read in contrapositive, now says: not continuous, therefore not differentiable at $(0,0)$.

So Step 2's claim is false, Step 4's claim is false, and Step 3's claim is false too: $f$ equals $\frac{1}{2}$ at points $\left(t,t^2\right)$ arbitrarily close to the origin, so “$f(x,y)\approx 0$ near $(0,0)$” is wrong — the plane $z=0$ misses the surface by $\frac{1}{2}$ no matter how close you zoom. Only Step 1 makes a true claim.

(c) From the definition. With $(\Delta x,\Delta y)=\left(t,t^2\right)$ and $t\ne 0$,

$$\Delta z=f\left(t,t^2\right)-f(0,0)=\frac{1}{2}-0=\frac{1}{2},$$

so the required identity would read

$$\frac{1}{2}=0\cdot t+0\cdot t^2+{\epsilon }_{1}t+{\epsilon }_2{t}^2.$$

If ${\epsilon }_1,{\epsilon }_2\to 0$ as $(\Delta x,\Delta y)\to (0,0)$, then as $t\to 0$ the right-hand side tends to $0\cdot 0+0\cdot 0=0$, while the left-hand side is the constant $\frac{1}{2}$. Contradiction; no such ${\epsilon }_1,{\epsilon }_2$ exist, and $f$ is not differentiable at the origin. The error in the tangent-plane approximation along this path is $\frac{1}{2}$ — not small compared with the distance moved, not even small — which is exactly what differentiability forbids.

(a) Case 1 of the chain rule. The tree has $z$ over $x$ and $y$, each over $t$: two paths, two terms,

$$\frac{dz}{dt}=f_x\frac{dx}{dt}+f_y\frac{dy}{dt}.$$

At $t=1$: $x=1+1=2$ and $y=3e^0=3$, so the partials are to be evaluated at $(2,3)$, which is where they were given. Also $\frac{dx}{dt}=2t=2$ and $\frac{dy}{dt}=3e^{t-1}=3$. Hence

$$\frac{dz}{dt}{|}_{t=1}=(-1)(2)+(4)(3)=10.$$

(b) Case 2, then square and add. The four inner partials are $u_x=2x$, $u_y=-2y$, $v_x=2y$, $v_y=2x$. Case 2 gives

$$w_x=f_u\cdot 2x+f_v\cdot 2y,w_y=f_u\cdot (-2y)+f_v\cdot 2x.$$

Squaring,

$$w_x^2=4x^2{f}_u^2+8xy{f}_u{f}_v+4y^2{f}_v^2,w_y^2=4y^2{f}_u^2-8xy{f}_u{f}_v+4x^2{f}_v^2.$$

The cross terms cancel, and collecting the rest gives

$$w_x^2+w_y^2=4\left(x^2+y^2\right)f_u^2+4\left(x^2+y^2\right)f_v^2=4\left(x^2+y^2\right)\left(f_u^2+f_v^2\right).$$

(c) Implicit differentiation formula. Let $F(x,y,z)=z^3+xz-y^2$. Then $F(3,2,1)=1+3-4=0$, so the point is on the surface; $F$ is a polynomial, so its partials are continuous; and

$$F_x=z,F_y=-2y,F_z=3z^2+x,$$

which at $(3,2,1)$ are $1$, $-4$ and $6$. Since $F_z=6\ne 0$, the formulas apply:

$$\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=-\frac{1}{6},\frac{\partial z}{\partial y}=-\frac{F_y}{F_z}=\frac{4}{6}=\frac{2}{3}.$$

The tangent plane to $z=z(x,y)$ at $(3,2,1)$ is

$$z-1=-\frac{1}{6}(x-3)+\frac{2}{3}(y-2).$$

Multiplying by $6$: $6z-6=-x+3+4y-8$, so $x-4y+6z=1$. Check: $3-8+6=1$, and the coefficients $⟨1,-4,6⟩$ are exactly $\nabla F(3,2,1)$, as they must be.

(a) Gradient, then a unit vector. $\nabla f=⟨e^y,x{e}^y-z^2,-2yz⟩$, so $\nabla f(P)=⟨1,1-4,0⟩=⟨1,-3,0⟩$. The vector from $P$ to $Q$ is $⟨2,-1,2⟩$, of length $\sqrt{4+1+4}=3$, so the unit direction is $\mathbf{u}=⟨\frac{2}{3},-\frac{1}{3},\frac{2}{3}⟩$ and

$$D_{\mathbf{u}}f(P)=⟨1,-3,0⟩\cdot ⟨\frac{2}{3},-\frac{1}{3},\frac{2}{3}⟩=\frac{2}{3}+1+0=\frac{5}{3}.$$

By the theorem on maximising the directional derivative, $f$ increases fastest in the direction of $\nabla f(P)=⟨1,-3,0⟩$ — as a unit vector, $\frac{1}{\sqrt{10}}⟨1,-3,0⟩$ — and the maximum rate is $|\nabla f(P)|=\sqrt{10}$. Sanity check: $\frac{5}{3}<\sqrt{10}$, as every directional derivative must be.

(b) Two known rates pin down the gradient. Both $\mathbf{u}$ and $\mathbf{v}$ are unit vectors ($\frac{9}{25}+\frac{16}{25}=1$) and they are perpendicular ($\frac{12}{25}-\frac{12}{25}=0$), so they form an orthonormal basis of the plane. Write $\nabla g(P)=\alpha \mathbf{u}+\beta \mathbf{v}$; then $D_{\mathbf{u}}g=\nabla g\cdot \mathbf{u}=\alpha$ and $D_{\mathbf{v}}g=\beta$, so $\alpha =3$, $\beta =-1$ and

$$\nabla g(P)=3⟨\frac{3}{5},\frac{4}{5}⟩-⟨\frac{4}{5},-\frac{3}{5}⟩=⟨\frac{9}{5}-\frac{4}{5},\frac{12}{5}+\frac{3}{5}⟩=⟨1,3⟩.$$

(Alternatively, solve the linear system $\frac{3}{5}a+\frac{4}{5}b=3$, $\frac{4}{5}a-\frac{3}{5}b=-1$ for $\nabla g=⟨a,b⟩$: multiplying by $3$ and $4$ and adding gives $5a=5$, then $b=3$.) Check: $⟨1,3⟩\cdot \mathbf{u}=\frac{3}{5}+\frac{12}{5}=3$ and $⟨1,3⟩\cdot \mathbf{v}=\frac{4}{5}-\frac{9}{5}=-1$.

The maximum rate of change is $|\nabla g(P)|=\sqrt{10}$, in the direction $\frac{1}{\sqrt{10}}⟨1,3⟩$. The rate is $0$ exactly when $\mathbf{w}\perp \nabla g(P)$, i.e. along the level curve: $\mathbf{w}=\pm \frac{1}{\sqrt{10}}⟨3,-1⟩$, since $⟨1,3⟩\cdot ⟨3,-1⟩=0$.

(c) The gradient is normal to level sets. The surface is the level set $F=6$ of $F(x,y,z)=xyz$, and by the theorem that the gradient is normal to level sets, $\nabla F(1,2,3)$ is a normal vector, provided it is nonzero:

$$\nabla F=⟨yz,xz,xy⟩,\nabla F(1,2,3)=⟨6,3,2⟩\ne 0.$$

Tangent plane: $6(x-1)+3(y-2)+2(z-3)=0$, that is $6x+3y+2z=18$; check $6+6+6=18$. Normal line, through $(1,2,3)$ in the direction $⟨6,3,2⟩$:

$$\frac{x-1}{6}=\frac{y-2}{3}=\frac{z-3}{2},\text{or}x=1+6t,y=2+3t,z=3+2t.$$

(a) Factor $f_x$, then split into cases. $f$ is a polynomial, so both partials exist everywhere and the critical points are exactly the simultaneous zeros of

$$f_x=2xy-2y=2y(x-1),f_y=x^2-2x+2y-3.$$

$f_x=0$ forces $y=0$ or $x=1$.

Case $y=0$. Then $f_y=x^2-2x-3=(x-3)(x+1)=0$, so $x=3$ or $x=-1$: the points $(3,0)$ and $(-1,0)$.

Case $x=1$. Then $f_y=1-2+2y-3=2y-4=0$, so $y=2$: the point $(1,2)$.

Three critical points: $(3,0)$, $(-1,0)$, $(1,2)$.

(b) Second Derivatives Test. $f_{xx}=2y$, $f_{xy}=2x-2$, $f_{yy}=2$, all polynomials, hence continuous on every disc — the hypothesis of the test. So

$$D=f_{xx}{f}_{yy}-f_{xy}^2=4y-4(x-1{)}^2.$$

At $(3,0)$: $D=0-16=-16<0$, a saddle point. At $(-1,0)$: $D=0-16=-16<0$, a saddle point. At $(1,2)$: $D=8-0=8>0$ and $f_{xx}=4>0$, a local minimum, with value

$$f(1,2)=2-4+4-6=-4.$$

(c) No — $f$ is unbounded below. Restrict $f$ to the line $y=-1$:

$$f(x,-1)=-x^2+2x+1+3=-x^2+2x+4⟶-\infty \text{as }x\to \infty ;$$

for instance $f(10,-1)=-100+20+4=-76<-4$. So no value is an absolute minimum, and $-4$ is a local minimum only. (Likewise $f(0,y)=y^2-3y\to \infty$, so there is no absolute maximum either.) There is no contradiction with the test: critical points are where a local extremum can be, and on an unbounded domain nothing — no Extreme Value Theorem — promises that an absolute extremum exists at all.

Existence. $f$ is a polynomial, hence continuous, and the closed disc is closed (it contains its boundary circle) and bounded (it lies inside the disc of radius $2$). By the Extreme Value Theorem for two variables, $f$ attains an absolute maximum and an absolute minimum on it.

(a) Interior. $f_x=2x$ and $f_y=-2y+2$ vanish together only at $(0,1)$, which is inside the disc since $0+1<4$; there $f(0,1)=-1+2=1$. The second partials $f_{xx}=2$, $f_{yy}=-2$, $f_{xy}=0$ are continuous and give $D=(2)(-2)-0=-4<0$: a saddle point. An absolute extremum attained at an interior point is in particular a local extremum there, and a saddle point is neither a local maximum nor a local minimum; so the value $1$ is out, and both absolute extrema must lie on the boundary circle.

(b) Boundary. On the circle $x^2=4-y^2$ with $-2\le y\le 2$, and because $f$ depends on $x$ only through $x^2$, the substitution is legitimate:

$$h(y)=\left(4-y^2\right)-y^2+2y=-2y^2+2y+4,-2\le y\le 2.$$

Closed-interval method: $h^{\prime }(y)=-4y+2=0$ at $y=\frac{1}{2}$, and

$$h\left(\frac{1}{2}\right)=-\frac{1}{2}+1+4=\frac{9}{2},h(2)=-8+4+4=0,h(-2)=-8-4+4=-8.$$

Translate back to points: $y=\frac{1}{2}$ gives $x^2=4-\frac{1}{4}=\frac{15}{4}$, so $x=\pm \frac{\sqrt{15}}{2}$; $y=2$ gives $(0,2)$; $y=-2$ gives $(0,-2)$. (Parametrising instead, $x=2\cos t$, $y=2\sin t$ gives $g(t)=4-8{\sin }^{2}t+4\sin t$, the same quadratic in $s=\sin t$, with its maximum at $\sin t=\frac{1}{4}$, i.e. $y=\frac{1}{2}$ — the substitution route simply avoids an ugly angle.)

Check one directly: $f\left(\frac{\sqrt{15}}{2},\frac{1}{2}\right)=\frac{15}{4}-\frac{1}{4}+1=\frac{14}{4}+1=\frac{9}{2}$.

(c) Compare. The candidates are $1$ (interior saddle, already excluded), $\frac{9}{2}$, $0$ and $-8$. The absolute maximum is $\frac{9}{2}$, at the two points $\left(\pm \frac{\sqrt{15}}{2},\frac{1}{2}\right)$; the absolute minimum is $-8$, at $(0,-2)$.

On the open disc the set is not closed, so the Extreme Value Theorem no longer applies, and in fact both extrema disappear: the only interior critical point is a saddle, so $f$ has no local extremum anywhere in the open disc, hence no absolute one. The values $\frac{9}{2}$ and $-8$ are approached — $f(0,y)=-y^2+2y\to -8$ as $y\to -2$ from inside — but never attained. The open disc has a supremum and an infimum of $f$, not a maximum and a minimum.

(a) Existence, gradient check, then cases. The ellipse is closed and bounded and $f$ is continuous, so by the Extreme Value Theorem both extremes exist. With $g(x,y)=x^2+2y^2$, $\nabla g=⟨2x,4y⟩$ vanishes only at the origin, which is not on the ellipse, so the method applies everywhere on it. Since $\nabla f=⟨2xy,x^2⟩$, the equations $\nabla f=\lambda \nabla g$ and $g=6$ are

$$2xy=2\lambda x(1),x^2=4\lambda y(2),x^2+2y^2=6(3).$$

Equation (1) says $x(y-\lambda )=0$, so either $x=0$ or $\lambda =y$.

Case $x=0$. Then (3) gives $2y^2=6$, $y=\pm \sqrt{3}$, and (2) reads $0=4\lambda y$, so $\lambda =0$, consistent. Candidates $(0,\pm \sqrt{3})$, where $f=0$.

Case $x\ne 0$, so $\lambda =y$. Then (2) gives $x^2=4y^2$, and (3) becomes $4y^2+2y^2=6$, so $y=\pm 1$ and $x=\pm 2$: four candidates $(\pm 2,1)$ and $(\pm 2,-1)$, with

$$f(\pm 2,1)=4,f(\pm 2,-1)=-4.$$

Comparing $0,4,-4$: the maximum is $4$ at $(2,1)$ and $(-2,1)$, the minimum is $-4$ at $(2,-1)$ and $(-2,-1)$. Dividing (1) through by $x$ would have silently discarded the first case; here those points happen not to be extremes, but nothing guaranteed that in advance.

(b) Two constraints. The curve is the intersection of the cylinder with a non-vertical plane, an ellipse: it is bounded, since $x^2+y^2=1$ bounds $x$ and $y$ and then $z=1-x-y$ is bounded too, and closed, so $z$ attains a maximum and a minimum on it. Take $f(x,y,z)=z$, $g=x^2+y^2$, $h=x+y+z$:

$$\nabla f=⟨0,0,1⟩,\nabla g=⟨2x,2y,0⟩,\nabla h=⟨1,1,1⟩.$$

These last two are never parallel on the curve (a multiple of $\nabla h$ has nonzero third component, while $\nabla g$ has third component $0$ and is itself nonzero on the cylinder), so the two-constraint condition $\nabla f=\lambda \nabla g+\mu \nabla h$ holds at each extremum:

$$0=2\lambda x+\mu ,0=2\lambda y+\mu ,1=\mu .$$

So $\mu =1$ and $2\lambda x=-1=2\lambda y$. This forces $\lambda \ne 0$, and then $x=y$. The cylinder gives $2x^2=1$, $x=y=\pm \frac{1}{\sqrt{2}}$, and the plane gives $z=1-x-y=1\mp \sqrt{2}$. The two candidates are

$$\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},1-\sqrt{2}\right)\text{and}\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},1+\sqrt{2}\right),$$

so the lowest point has $z=1-\sqrt{2}$ and the highest has $z=1+\sqrt{2}$. Check by parametrising: $x=\cos t$, $y=\sin t$ gives $z=1-\cos t-\sin t=1-\sqrt{2}\sin \left(t+\frac{\pi }{4}\right)$, whose range is exactly $\left[1-\sqrt{2},1+\sqrt{2}\right]$.