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MATH 180\hfill FALL 1995

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%10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#30}

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5.2, \#17.
The behavior of $f(x)=e^{1/x}$ depends on the behavior of $1/x$
at the given limits.  We find
$$
  \eqalign{&x\rightarrow\infty\cr &x\rightarrow-\infty\cr
           &x\rightarrow0^+\cr &x\rightarrow0^-\cr}\quad
  \eqalign{&1/x\rightarrow0\cr &1/x\rightarrow0\cr
           &1/x\rightarrow\infty\cr &1/x\rightarrow-\infty\cr}\quad
  \eqalign{&f(x)\rightarrow1\cr &f(x)\rightarrow1\cr
           &f(x)\rightarrow\infty\cr &f(x)\rightarrow0\cr}$$
$\displaystyle f'(x)=-{1\over x^2}e^{1/x},\quad
f''(x)={1+2x\over x^4}e^{1/x}$.  Since
$f'(x)$ is always negative, $f$ is decreasing on its domain.
$f''(x)$ is zero at $x=-1/2$, $f''(x)$ is negative if $x<-1/2$, and
$f''(x)$ is positive if $x>-1/2$ and $x\ne0$.
(The point $x=0$ is not in the domain of $f$).
So $f$ is concave down on $(-\infty,-1/2)$ and concave-up on
$(-1/2,0)$ and $(0,\infty)$.

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5.2, \#23.  The graph should that of a decreasing function which
alternates between concave-up and concave down in the four intervals
determined by the points $x_1$, $x_2$, $x_3$.

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5.3, \#2.  $p(x)=x^3-ax$, $p'(x)=3x^2-a$, $p''(x)=6x$.
If $a<0$, then $p'(x)=3x^2-a$ is always positive and
$p(x)$ is always increasing.  If $a>0$, then $p'(x)=3x^2-a$ has roots
at $\pm\sqrt{a/3}$.  Since $p'$ is positive for $x<-\sqrt{a/3}$,
negative for $-\sqrt{a/3}<x<\sqrt{a/3}$, and positive
for $x>\sqrt{a/3}$, it follows $p$ has a local maximum at
$x=-\sqrt{a/3}$, a local minimum at $x=\sqrt{a/3}$.
$p''(x)$ changes sign at $x=0$.  So
$p$ is concave-down for $x<0$ and concave-up for $x>0$.
The graphs are the cubics (a) and (c) illustrated in figure 5.34
on page 265 of the text.

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5.3, \#3.  The discussion in the preceding problem shows $p$ has
a local maximum at $x=-\sqrt{a/3}$ and a local minimum at
$x=\sqrt{a/3}$.  Now
$$f(-\sqrt{a/3})=(2a/3)\sqrt{a/3}),\quad
  f(\sqrt{a/3})=-(2a/3)\sqrt{a/3}).$$
As $a$ increases, the critical point $(-\sqrt{a/3},(2a/3)\sqrt{a/3})$
moves left and up,
the critical point $(\sqrt{a/3},-(2a/3)\sqrt{a/3})$
moves right and down.

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5.3, \#4.  $f(x)=x^2+2ax$, $f'(x)=2x+2a$.  The one critical point
$(-a, -a^2)$ of $f$ is a local minimum.  Indeed,
$(-a, -a^2)$ is the vertex of the parabola $f(x)$.
The roots of $f(x)$ are $x=-2a$ and $x=0$ (this is easily ssen by
factoring $f(x)=x(x+2a)$).
As $a$ increases, the vertex $(-a, -a^2)$
moves from right to left along the graph of $y=-x^2$;
the root $x=-2a$ moves from right to left;
the root $x=0$ stays fixed.

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