Chapter 9 Functions of Several Variables
Exercise 13
(By analambanomenos) Since $\big|\mathbf f(t)|^2=1$ is constant, we have
\[
0=\frac{d}{dt}\big|\mathbf f(t)\big|^2=\frac{d}{dt}\sum_{i=1}^3f_i^2(t)=2\sum_{i=1}^3f_i’(t)f_i(t)=2\mathbf f’(t)\cdot\mathbf f(t).
\]So $\mathbf f’(t)$ is perpendicular to $\mathbf f(t)$. I suppose this hasn’t been shown, but if $\mathbf a\cdot\mathbf b=0$, then $$|\mathbf a+\mathbf b|^2=|\mathbf a|^2+|\mathbf b|^2$$ and you can apply the law of cosines to the triangle formed by $\mathbf a$, $\mathbf b$, and $\mathbf a+\mathbf b$ to conclude that $\mathbf a$ and $\mathbf b$ form a right angle. Also, the book hasn’t mentioned tangent vectors, but this says that tangent vectors of curves on a sphere are perpendicular to the radial vectors, or that the tangent plane at the point on the sphere is perpendicular to the radius.
Exercise 14
(By analambanomenos) This function is easier to visualize if we convert to polar coordinates, $x=r\cos\theta$, $y=r\sin\theta$, with $r\ge0$ and $0\le\theta<2\pi$. Then we have $f(r,\theta)=r\cos^3\theta$, so that $f$ is linear in $r$ along rays from the origin, and since $\cos^3(\theta+\pi)=-\cos^3\theta$, $f$ is a linear function along every line through the origin.
(a) Keeping in mind that along the $x$ axis $f(x,0)=x$, and along the $y$ axis $f(0,y)=0$, we have
\begin{align*}
D_1f(x,y) &=
\begin{cases}
\displaystyle\frac{x^2(x^2+3y^2)}{(x^2+y^2)^2}, & (x,y)\ne(0,0) \\
1, & (x,y)=(0,0)
\end{cases} \\
D_2f(x,y) &=
\begin{cases}
\displaystyle\frac{2x^3y}{(x^2+y^2)^2}, & (x,y)\ne(0,0) \\
0, & (x,y)=(0,0)
\end{cases}
\end{align*}Converting to polar coordinates and simplifying we have, for $r>0$ and away from the origin,
\begin{align*}
D_1f(r,\theta) &= \cos^2\theta(\cos^2\theta+3\sin^2\theta) \\
D_2f(r,\theta) &= -2\cos^3\theta\sin\theta.
\end{align*}These do not depend on $r$ and are continuous functions of $\theta$, which attain their maximum and minimum values on the compact set $[0,2\theta]$. Excluding the origin, $D_1f$ and $D_2f$ take these maximum and minimum values along certain rays emanating from the origin. Hence they are bounded on $\mathbf R^2$.
(b) We can express $\mathbf u$ as $\cos\theta\mathbf e_1+\sin\theta\mathbf e_2$ for some $0\le\theta<2\pi$. Then, from the above expression of $f$ in polar coordinates, we have $f(t\mathbf u)=t\cos^3\theta$ for all $t\in\R1$ so that $D_{\mathbf u}f(0,0)=\cos^3\theta$, whose absolute value is less than 1.
(c) Since $f$ is differentiable away from the origin, we only need to show that $g’(0)$ exists. Since $\gamma(t)$ is differentiable at $t=0$, there are functions $\delta_1(t)$ and $\delta_2(t)$ such that, for $j=1,2$, $\gamma_j(t)=t\gamma_j’(0)+t\delta_j(t)$ and $\lim\delta_j(t)=0$ as $t\rightarrow0$. Hence
\begin{align*}
g’(0) &= \lim_{t\rightarrow0} \frac{g(t)}{t} \\
&= \lim_{t\rightarrow0} \frac{f\big(\gamma_1(t),\gamma_2(t)\big)}{t} \\
&= \lim_{t\rightarrow0} \frac{f\big(t\gamma_1′(0)+t\delta_1(t),t\gamma_2′(0)+t\delta_2(t)\big)}{t} \\
&= \lim_{t\rightarrow0} \frac{1}{t}\cdot\frac{t^3\big(\gamma_1′(0)+\delta_1(t)\big)^3}{t^2\Big(\big(\gamma_1′(0)+\delta_1(t)\big)^2+\big(\gamma_2′(0)+\delta_2(t)\big)^2\Big)} \\
&= \frac{\gamma_1′(0)^3}{\big|\gamma’(0)\big|^2}
\end{align*}Also, if $\gamma_r(t)$ and $\gamma_\theta(t)$ are the polar coordinates of $\gamma(t)$ for $t\ne 0$, then we have
\begin{align*}
\lim_{t\rightarrow0}\cos\gamma_\theta(t) &= \lim_{t\rightarrow0}\frac{\gamma_1(t)}{\big|\gamma(t)\big|} \\
&= \lim_{t\rightarrow0}\frac{t\big(\gamma’(0)+\delta_1(t)\big)}{t\sqrt{\big(\gamma_1′(0)+\delta_1(0)\big)^2+\big(\gamma_2′(0)+\delta_2(0)\big)^2}} \\
&= \frac{\gamma_1′(0)}{\big|\gamma’(0)\big|} \\
\lim_{t\rightarrow0}\sin\gamma_\theta(t) &= \lim_{t\rightarrow0}\frac{\gamma_2(t)}{\big|\gamma(t)\big|} \\
&= \lim_{t\rightarrow0}\frac{t\big(\gamma’(0)+\delta_2(t)\big)}{t\sqrt{\big(\gamma_1′(0)+\delta_1(0)\big)^2+\big(\gamma_2′(0)+\delta_2(0)\big)^2}} \\
&= \frac{\gamma_2′(0)}{\big|\gamma’(0)\big|}.
\end{align*} For $t$ where $\gamma(t)\ne\mathbf 0$, we can apply the special case of the chain rule, Theorem 9.15, worked out in Example 9.18 to get
\begin{align*}
g’(t) &= D_1f\big(\gamma_1(t),\gamma_2(t)\big)\gamma_1′(t)+D_2f\big(\gamma_1(t),\gamma_2(t)\big)\gamma_2′(t) \\
&= \cos^2\gamma_\theta(t)\big(\cos^2\gamma_\theta(t)+3\sin^2\gamma_\theta(t)\big)\gamma_1′(t)-2\big(\cos^3\gamma_\theta(t)\sin\gamma_\theta(t)\big)\gamma_2′(t)
\end{align*}Now assume that $\gamma\in\mathscr C’$. Applying the above limits we get
\begin{align*}
\lim_{t\rightarrow0}g’(t) &= \frac{\gamma_1′(0)^2}{\big|\gamma’(0)\big|^2}\Bigg(\frac{\gamma_1′(0)^2}{\big|\gamma’(0)\big|^2}+3\frac{\gamma_2′(0)^2}{\big|\gamma’(0)\big|^2}\Bigg)\gamma_1′(0)-
2\Bigg(\frac{\gamma_1′(0)^3}{\big|\gamma’(0)|^3}\frac{\gamma_2′(0)}{\big|\gamma’(0)\big|}\Bigg)\gamma_2′(0) \\
&= \frac{\gamma_1′(0)^3\big(\gamma_1′(0)^2+\gamma_2′(0)^2\big)}{\big|\gamma’(0)\big|^4} \\
&= \frac{\gamma_1′(0)^3}{\big|\gamma’(0)\big|^2} \\
&= g’(0).
\end{align*}Hence $g$ is continuous at 0, and so $g\in\mathscr C’$.
(d) Let $\mathbf u=\big(\cos(\pi/4),\sin(\pi/4)\big)$, a unit vector, and suppose $f$ were differentiable at $(0,0)$. Then by formula (40) in Example 9.18 in the text, the directional derivative
would be
\[
D_{\mathbf u}f(0,0)=D_1(0,0)\cos(\pi/4)+D_2(0,0)\sin(\pi/4)=\cos(\pi/4)=\sqrt{2}/2.
\]However, by part (b), $D_{\mathbf u}f(0,0)=\cos^3(\pi/4)=\sqrt{2}/4$. Hence $f$ cannot be differentiable at $(0,0)$.
Exercise 15
(By analambanomenos)
(a) We have
\[
(x^4+y^2)^2-4x^4y^2 = x^8-2x^4y^2+y^4-4x^4y^2 = x^8-2x^4y^2+y^4=(x^4-y^2)^2 \ge 0
\]so that $4x^4y^2\le(x^4+y^2)^2$. To show that $f$ is continuous, it suffices to show this at $(0,0)$, since $f$ is differentiable otherwise. We have
\[
\big|f(x,y)\big| \le x^2+y^2+2x^2|y|+x^2\frac{4x^4y^2}{(x^4+y^2)^2} \le 2x^2+y^2+2x^2|y|
\]which goes to $f(x,y)=0$ as $(x,y)\rightarrow 0$.
(b) We have $g_\theta(0)=f(0,0)=0$, and for $t\ne 0$,
\begin{align*}
g_\theta(t) &= t^2\cos^2\theta + t^2\sin^2\theta-2t^3\cos^2\theta\sin\theta – \frac{4t^8\cos^6\theta\sin^2\theta}{(t^4\cos^4\theta+t^2\sin^2\theta)^2} \\
&= t^2 – 2t^3\cos^2\theta\sin\theta – \frac{4t^4\cos^6\theta\sin^2\theta}{(t^2\cos^4\theta+\sin^2\theta)^2}
\end{align*}The denominator in the last term is nonzero unless $\theta=0$ and $t=0$. Excluding the case $\theta=0$ this expression for $g_\theta(t)$ is true for all values of $t$ including 0, so we can differentiate it using the usual rules of calculus to get
\begin{align*}
g_\theta’(t) &= 2t – 6t^2\cos^2\theta\sin\theta – \frac{16t^3\cos^6\theta\sin^4\theta}{(t^2\cos^2\theta+\sin^2\theta)^3} \\
g_\theta”(t) &= 2 – 12t\cos^2\theta\sin\theta + \frac{48t^2\cos^6\theta\sin^4\theta(t^2\cos^2\theta-\sin^2\theta)}{(t^2\cos^2\theta+\sin^2\theta)^4}
\end{align*}Hence, for $\theta\ne 0$ we have $g_\theta’(0)=0$ and $g_\theta”(0)=2$. For the special case $\theta=0$, the formula for $g_\theta(t)$ simplifies to $g_0(t)=t^2$, so that we also have $g_0′(0)=0$ and $g_0”(0)=2$.
(c) We have
\[
f(x,x^2) = x^2+x^4-2x^2x^2-\frac{4x^6x^4}{(x^4+x^4)^2} = x^2+x^4-2x^4-x^2 = -x^4.
\]
Exercise 16
(By analambanomenos) The derivative of $f$ at 0 is
\[
f’(0)=\lim_{t\rightarrow 0}\frac{f(t)}{t}=\lim_{t\rightarrow 0}\big(1+2t\sin(t^{-1})\big)=1 \\
\]since $\big|2t\sin(t^{-1})\big|\le 2t\rightarrow 0$ as $t\rightarrow 0$.
For $t\in(-1,1)$, $t\ne 0$, we have
\begin{align*}
f’(t) &= 1+4t\sin(t^{-1})-2\cos(t^{-1}) \\
\big|f’(t)\big| &\le 1+4\big|t\sin(t^{-1})\big|+2\big|\cos(t^{-1})\big| \\
&\le 3+4|t| \\
&\le 7
\end{align*}If $n$ is a positive integer, let $a_n=2n\pi$, $b_n=(2n+1)\pi$. Then, since $\sin a_n=\sin b_n=0$, $\cos a_n=1$, and $\cos b_n=-1$, we have $f’(a_n^{-1})=-1$ and $f’(b_n^{-1})=3$. Hence $f$ has a local maximum in $(b_n^{-1},a_n^{-1})$ and a local minimum in $(a_n^{-1},b_{n+1}^{-1})$ and so cannot be one-to-one in either interval. Since a neighborhood of 0 contains an infinite number of such intervals, $f$ cannot be one-to-one in any neighborhood of 0.
Exercise 17
(By analambanomenos)
(d) Fix $y=y_0$. Then $\mathbf f$ maps $(x,y_0)$, $x\in\mathbf R$, to the ray from (but not including) the origin through the point $(\cos y_0,\sin y_0)$ on the unit circle. The negative values of $x$ map to points on the ray inside the circle, and the positive values of $x$ map to points outside the unit circle.
Fix $x=x_0$. Then $\mathbf f$ maps $(x_0,y)$, $v\in\mathbf R$, to the circle of radius $e^x$. Note that $$\mathbf f(x_0,y)=\mathbf f(x_0,y+2n\pi)$$ for all integers $n$.
(a) The results of (d) show that the range of $\mathbf f$ is $\R2$ minus the origin.
(b) Since
\begin{align*}
D_1f_1(x,y) = \phantom{-}e^x\cos y &\qquad D_1f_2(x,y) = e^x\sin y \\
D_2f_1(x,y) = -e^x\sin y &\qquad D_1f_2(x,y) = e^x\cos y
\end{align*}the Jacobian of $\mathbf f$ at $(x,y)$ is $e^{2x}(\cos^2y+\sin^2y)=e^{2x}\ne 0$ for all $x$. From (d) we have $\mathbf f(x,y)=\mathbf f(x,y+2n\pi)$ for all integers $n$, so $\mathbf f$ is not one-to-one on $\R2$.
(c) Let $w=f_1(x,y)=e^x\cos y$, $z=f_2(x,y)=e^x\sin y$. Then, in a neighborhood of $\mathbf b=(1/2,\sqrt{3}/2)$ where $w\ne 0$, we have $w^2+z^2=e^{2x}$, $z/w=\tan y$, so
\begin{gather*}
\mathbf g(w,z)=\big(\log\sqrt{w^2+z^2},\arctan(z/w)\big) \\
\mathbf f’(x,y)=
\begin{pmatrix}
\phantom{-}e^x\cos y & e^x\sin y \\
-e^x\sin y & e^x\cos y
\end{pmatrix} \\
\mathbf g’(w,z)=
\begin{pmatrix}
w/(w^2+z^2) & -z/(w^2+z^2) \\
z/(w^2+z^2) & \phantom{-}w/(w^2+z^2)
\end{pmatrix} \\
\mathbf f’(\mathbf a)\circ\mathbf g’(\mathbf b)=
\begin{pmatrix}
\phantom{-}1/2 & \sqrt{3}/2 \\
-\sqrt{3}/2 & 1/2
\end{pmatrix}
\begin{pmatrix}
1/2 & -\sqrt{3}/2 \\
\sqrt{3}/2 & \phantom{-}1/2
\end{pmatrix}
\=
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\end{gather*}
Exercise 18
(By analambanomenos) Let the mapping be denoted by $\mathbf F=(f_1,f_2)$. Then $\mathbf F$ maps the points $(x,0)$ and $(-x,0)$ on the $x$-axis to $(x^2,0)$ on the positive $u$-axis, and for a fixed $y_0\ne 0$, $\mathbf F$ maps the lines $(x,y_0)$ and $(x,-y_0)$ to the parabola $u=v^2/(4y_0)^2-y_0^2$, which is symmetric with respect to the $u$-axis and passes through and otherwise lies to the left of the point $(-y_0^2,0)$. If $y_0>0$, then the points $(x,y_0)$, $x>0$, are mapped to the upper branch of the parabola, and the points $(x,y_0)$, $x<0$, are mapped to the lower branch, and the opposite is true if $y_0<0$.
Similarly, $\mathbf F$ maps the point $(0,y)$ and $(0,-y)$ on the $y$-axis to $(-y^2,0)$ on the $u$-axis, and for a fixed $x_0\ne 0$, $\mathbf F$ maps the lines $(x_0,y)$ and $(-x_0,y)$ to the parabola $u=-v^2/(4x_0)^2+ x_0^2$, which is symmetric with respect to the $u$-axis and passes through and otherwise lies to the right of the point $(x_0^2,0)$. If $x_0>0$, then the points $(x_0,y)$, $y>0$, are mapped to the upper branch of the parabola, and the points $(x_0,y)$, $y<0$, are mapped to the lower branch, and the opposite is true if $x_0<0$.
Other than mapping the origin in the $x$-$y$ plane to the origin in the $u$-$v$ plane, $\mathbf F$ maps two distinct points in the $x$-$y$ plane to each point in the $u$-$v$ plane outside the origin. If we let $w= \sqrt{u^2+v^2}$ be the distance from the origin to the point $(u,v)$, then
\[
\Bigg(\sqrt{\frac{w+u}{2}},\sqrt{\frac{w-u}{2}}\Bigg)\quad\hbox{and}\quad\Bigg(-\sqrt{\frac{w+u}{2}},-\sqrt{\frac{w-u}{2}}\Bigg)
\]are mapped by $\mathbf F$ to the point $(u,v)$ if $v$ is positive, and
\[
\Bigg(\sqrt{\frac{w+u}{2}},-\sqrt{\frac{w-u}{2}}\Bigg)\quad\hbox{and}\quad\Bigg(-\sqrt{\frac{w+u}{2}},\sqrt{\frac{w-u}{2}}\Bigg)
\]are mapped by $\mathbf F$ to the point $(u,v)$ if $v$ is negative.
Since
\begin{align*}
D_1f_1(x,y) = 2x &\qquad D_2f_1(x,y) = -2y \\
D_1f_2(x,y) = 2y &\qquad D_2f_2(x,y) = 2x
\end{align*}the Jacobian of $\mathbf f$ at $(x,y)$ is $4(x^2+y^2)$ which is nonzero except at the origin. If we exclude the origin in both planes, then $\mathbf F$ is locally one-to-one, but globally two-to-one.
Letting $\mathbf b = (3,4)$, then $\mathbf F$ maps the point $\mathbf a=(2,1)$ to $\mathbf b$. Again letting $w= \sqrt{u^2+v^2}$ be the distance from the origin to the point $(u,v)$, locally $\mathbf F$ has the inverse function
\begin{gather*}
\mathbf G(u,v)= \Bigg(\sqrt{\frac{w+u}{2}},\sqrt{\frac{w-u}{2}}\Bigg) \\
\mathbf F’(x,y)=
\begin{pmatrix}
2x & -2y \\
2y & \phantom{-}2x
\end{pmatrix} \\
\mathbf G’(u,v)=
\begin{pmatrix}
\frac{u+w}{4w}\sqrt{\frac{2}{w+u}} & \frac{v}{4w}\sqrt{\frac{2}{w+u}} \\
\frac{u-w}{4w}\sqrt{\frac{2}{w-u}} & \frac{v}{4w}\sqrt{\frac{2}{w-u}}
\end{pmatrix} \\
\mathbf F’(\mathbf a)\circ\mathbf G’(\mathbf b)=
\begin{pmatrix}
4 & -2 \\
2 & \phantom{-}4
\end{pmatrix}
\begin{pmatrix}
\phantom{-}1/5 & 1/10 \\
-1/10 & 1/5
\end{pmatrix}
\=
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\end{gather*}
Exercise 19
(By analambanomenos) Let
\begin{align*}
\mathbf f(x,y,z,u) &= \big(f_1(x,y,z,u),f_2(x,y,z,u),f_3(x,y,z,u)\big) \\
&= (3x+y-z+u^2,x-y+2z+u,2x+2y-3z+2u.
\end{align*}Then the matrix of $\mathbf f’(x,y,z,u)$ is
\[
\big(D_jf_i(x,y,z,u)\big)=
\begin{pmatrix}
3 & \phantom{-}1 & -1 & 2u \\
1 & -1 & \phantom{-}2 & 1 \\
2 & \phantom{-}2 & -3 & 2
\end{pmatrix}.
\]Note that $\mathbf f(\mathbf 0)=\mathbf 0$. The $x,y,u$ part of $\mathbf f’$ has determinant $8u-12\ne 0$ near $\mathbf 0$, so by the implicit function theorem, there is a solution of $\mathbf f\big(x(z),y(z),z,u(z)\big)=\mathbf 0$ near $\mathbf 0$.
Similarly, the determinant of the $x,z,u$ part of $\mathbf f’$ is equal to $21-14u\ne 0$ near $\mathbf 0$, so there is a solution of $\mathbf f\big(x(y),y,z(y),u(y)\big)=\mathbf 0$ near $\mathbf 0$. And the determinant of the $y,z,u$ part of $\mathbf f’$ is equal to $3-2u\ne 0$ near $\mathbf 0$, so there is a solution of $\mathbf f\big(x,y(x),z(x),u(x)\big)=\mathbf 0$ near $\mathbf 0$.
However, the determinant of the $x,y,z$ part of $\mathbf f’$ is equal to 0, so the implicit function theorem cannot be applied. If you try to solve for $x,y,z$ in terms of $u$, you just get an equation in $u$ which has no solution near $\mathbf 0$.
Exercise 20
(By analambanomenos) If the real-valued function $f(x,y)$ is smooth and nonconstant in a region of $\mathbf R^2$, then the solution of $f(x,y)=0$ is locally a smooth curve. If $D_1f(x_0,y_0)\ne 0$ at a point of the curve, then the curve doesn’t have a vertical tangent at $(x_0,y_0)$, and so it will be the graph of a function $y=g(x)$ near $x=x_0$, so that $f\big(x,g(x)\big)=0$. Similarly, if $D_2f(x_0,y_0)\ne 0$, then the curve doesn’t have a horizontal tangent at $(x_0,y_0)$, and so it will be the graph of a function $x=h(y)$ near $y=y_0$, so that $f\big(h(y),y\big)=0$.
Exercise 21
(By analambanomenos)
(a) We have $D_1f(x,y)=6x(x-1)$ so $D_1f(x,y)$ if $x=0$ or $x=1$. Also, $D_2f(x,y)=6y(6+1)$ so $D_2f(x,y)=0$ if $y=0$ or $y=-1$. Hence the gradient of $f$ equals 0 at the four points $(0,0)$,
$(1,0)$, $(0,-1)$, and $(1,-1)$. To tell whether these are local maxima or minima, or saddle points, the easiest way is to apply the second derivative test for multivariable functions, which involves finding the eigenvalues of the matrix of second derivatives (the “Hessian”) at these points, but that wasn’t demonstrated in the text, so we need to show this more directly.
At $(1,0)$, for small values $d_1$ and $d_2$, we have
\[
f(1+d_1,d_2)-f(1,0)=d_1^2(2d_1+3)+d_2^2(2d_2+3)
\]which is positive for small values of $d_1$ and $d_2$. Hence $f$ has a local minimum at $(1,0)$.
Similarly, at $(0,-1)$ we have
\[
f(d_1,-1+d_2)-f(0,-1)=d_1^2(2d_1-3)+d_2^2(2d_2-3)
\]which is negative for small values of $d_1$ and $d_2$. Hence $f$ has a local maximum at $(1,0)$.
At $(0,0)$ $f(x,0)=2x^3-3x^2$ can be shown to have a local maximum at $x=0$, using the usual calculus techniques. However, $f(0,y)=2y^3+3y^2$ has a local minimum at $y=0$. Hence $(0,0)$ is a saddle point for $f$.
Similarly, at $(1,-1)$ $f(x,-1)=2x^3-3x^2+1$ has a local minimum at $x=1$, but $$f(1,y)=2y^3+3y^2-1$$ has a local maximum at $y=-1$. Hence $(1,-1)$ is also a saddle point for $f$.
(b) Note that
\[
f(x,y)=(x+y)\big(2(x^2-xy+y^2)-3(x-y)\big).
\] The first factor shows that $f$ is equal to zero along the diagonal $y=-x$. The zero set of the second factor of degree 2 must be some sort of conic section. Converting to polar coordinates we get
\[
r(\theta)=\frac{3}{2}\bigg(\frac{\cos\theta-\sin\theta}{1-\cos\theta\sin\theta}\bigg)
\]which can be seen to describe a ellipse symmetric with the diagonal line $y=-x$ and intersecting it at the points $(0,0)$ and $(1,-1)$.
The points of the zero set along the diagonal satisfy the relation $y=-x$ except at the two intersection points, where we cannot describe $x$ or $y$ as single-valued functions of each other.
Let $g(x,y)=2(x^2-xy+y^2)-3(x-y)$. Along the zero set of $g(x,y)$, $x$ cannot be expressed as a function of $y$ where $D_1g(x,y)=4x-2x-3=0$. (I am using the elliptical zero set as described above; the general case is more complicated.) That is, we are looking for the intersection of the zero set of $g$ with the line $y=2x-(3/2)$, which occurs at the points $(0,-3/2)$ and $(1,1/2)$.
Similarly, $y$ cannot be expressed as a function of $x$ where $D_2g(x,y)=-2x+4y+3=0$, so we are looking for the intersection of the zero set of $g$ with the line $x=2y+(3/2)$, which occurs at the points $(3/2,0)$ and $(-1/2,-1)$.
Exercise 22
(By analambanomenos) We have
\begin{align*}
D_1f(x,y) &= 6(x^2+y^2-x) \\
D_2f(x,y) &= (12x+6)y
\end{align*}so $D_2f(x,y)=0$ if $y=0$ or $x=-1/2$. If $y=0$, then $D_1f(x,y)=6x(x-1)=0$ if $x=0$ or $x=1$. However, if $x=-1/2$, then $D_1f(x,y)=6(3/4+y^2)>0$ for all $y$. Hence the gradient of $f$ equals $\mathbf 0$ only at $(0,0)$ and $(1,0)$.
At $(1,0)$, let $d_1,d_2$ be near 0. Then
\begin{align*}
f(1+d_1,d_2) – f(1,0) &= 2(1+d_1)^3+6(1+d_1)d_2^2-3(1+d_1)^2+3d_2^2+1 \\
&= (d_1^2+3d_2^2)(2d_1+3)
\end{align*}which is positive for small values of $d_1$ and $d_2$, hence $f$ has a local minimum at $(1,0)$.
At $(0,0)$, the values of $f$ along the $x$-axis, $f(x,0)=2x^3-3x^2$, have a local maximum at $x=0$, while the values of $f$ along $y$-axis, $f(0,y)=3y^2$ have a local minimum at $y=0$. Hence $f$ has a saddle point at $(0,0)$.
Solving $f(x,y)=0$ for $y$, we get
\[
y=\pm x\sqrt{\frac{3-2x}{3(2x+1)}}
\]
The graph of this looks like the folium of Descartes, only with a vertical asymptote of $x=-1/2$ and symmetrical with the $x$-axis, with a double point at the origin and intersecting the $x$-axis at 0 and 3/2.
At the origin, a double-point, we cannot solve for $x$ in terms of $y$, or vice versa. Otherwise we can solve for $x$ in terms of $y$ except where $D_2f(x,y)=6y(2x+1)=0$, which on the zero set
only occurs at the $x$-axis intercepts of the origin and $(3/2,0)$. We can solve for $y$ in terms of $x$ away from the origin except where $D_1f(x)=6(x^2+y^2-x)=0$. Inserting the expression for $y$ above in this equation and solving for $x$, we see that this happens only at origin and the points
\[
\bigg(\frac{\sqrt{3}}{2},\frac{1}{2}\sqrt{2\sqrt{3}-3}\bigg)\quad\hbox{and}\quad\bigg(\frac{\sqrt{3}}{2},-\frac{1}{2}\sqrt{2\sqrt{3}-3}\bigg)
\]
