HW8 Solutions

I skipped writing the solution for many problems. But if you need
any that I omitted, or even a problem that you tried apart from the
homework set, let me know. I'll get around to it.

Homework 8

4.3: 6, 8, 14, 20, 22, 30, 50, 68, 82

4.4: 2, 6, 12, 14, 18, 42, 52, 54, 64, 80

4.7: 24, 50



4.3:

\#6

$f$ is increasing wherever $f^{\prime}>0$ and decreasing wherever
$f^{\prime}<0$. Thus $f$ increasing on $[0,1)\cup(3,5)$ and decreasing on $(1,3)\cup(5,6)$. And so $f$ has a local maximum at $x=1$ and $x=5$ and local minimum at $x=3$. \#8 \#14 \#20 \#22 $f^{\prime}(x)=4x^{3}(x-1)^{3}+x^{4}\left(3(x-1)^{2}\right)=x^{3}(x-1)^{2}\left(4(x-1)+3x\right)=x^{3}(x-1)^{2}(7x-4)$ The derivative has zeros at $x=0$, $x=1$, and $x=\frac{4}{7}$. $f^{\prime\prime}(x)=3x^{2}(x-1)^{2}(7x-4)+2(x-1)x^{3}(7x-4)+7x^{3}(x-1)^{2}$ It's clear that $f^{\prime\prime}(0)=0$, $f^{\prime\prime}(1)=0$, and $f^{\prime\prime}(\frac{4}{7})=7\left(\frac{4}{7}\right)^{3}\left(\frac{4}{7}-1\right)^{2}>0$

This the second derivative test is uninformative at $x=0$ and $x=1$.

So we should evaluate $f^{\prime}$ at points near $0$ and $1$.
It is convenient to write $(7x-4)$ as $7(x-\frac{4}{7})$.

Then without fulling evaluating we have $f^{\prime}(-1)=7(-1)^{3}(-1-1)^{2}(-1-\frac{4}{7})>0$

$f^{\prime}(\frac{1}{2})=7(\frac{1}{2})^{3}(\frac{1}{2}-1)^{2}(\frac{1}{2}-\frac{4}{7})<0$ $f^{\prime}(\frac{3}{2})=7(\frac{3}{2})^{3}(\frac{3}{2}-1)^{2}(\frac{3}{2}-\frac{4}{7})>0$

$f^{\prime}(2)=7(2)^{3}(2-1)^{2}(2-\frac{4}{7})>0$

Where $-1<0<\frac{1}{2}<\frac{4}{7}<\frac{3}{2}<1<2$. \#30 \#50 $f^{\prime}(x)=\frac{e^{x}(1+e^{x})-e^{x}e^{x}}{\left(1+e^{x}\right)^{2}}=\frac{e^{x}}{\left(1+e^{x}\right)^{2}}>0$

$f^{\prime\prime}(x)=\frac{e^{x}(1+e^{x})^{2}-2(1+e^{x})e^{x}e^{x}}{\left(1+e^{x}\right)^{4}}=\frac{e^{x}(1+e^{x})-2e^{x}e^{x}}{\left(1+e^{x}\right)^{3}}=\frac{e^{x}(1-e^{x})}{\left(1+e^{x}\right)^{3}}$

Then the sign of $f^{\prime\prime}(x)$ depends on the factor $(1-e^{x})$.
It is positive for $x<0$ and negative for $x>0$.

\#68

One needs to compute the first and second derivatives. If there is
to be a maximum, the first derivative must equals 0 at $x=2$. Also,
we must have $f(2)=1$which implies that $a2e^{b4}=1$. These two
conditions restricts the possible values for $a$ and $b$.

$f^{\prime}(x)=ae^{bx^{2}}+axe^{bx^{2}}(2bx)$.

Then $0=f^{\prime}(2)=ae^{b4}+a2e^{b4}(2b2)=\frac{1}{2}+1(4b)=\frac{1}{2}+4b$.
This implies that $b=-\frac{1}{8}$.

Then we have $a=\frac{1}{2}e^{1/2}$.

I suppose we could consider checking that our answer does have a maximum.

$f^{\prime}(x)=ae^{bx^{2}}(1+2bx^{2})=ae^{bx^{2}}(1-\frac{1}{4}x^{2})$.

$f^{\prime\prime}(x)=ae^{bx^{2}}(-\frac{1}{2}x+2bx)=ae^{bx^{2}}(-\frac{3}{4}x)$.
So that $f^{\prime\prime}(2)<0$. Jolly good. \#82 With practice, one can start from a graph of $f^{\prime\prime\prime}$ and infer a graph of $f^{\prime\prime}$ then $f^{\prime}$ and then $f$. In general for higher derivatives. $f^{\prime\prime\prime}(c)>0$ implies $f^{\prime\prime}(x)$ is increasing
around $c$. Because we know $f^{\prime\prime}(c)=0$ then $f^{\prime\prime}$
is negative for $x<c$ and positive for $x>c$.

Thus $f^{\prime}$is decreasing for $x<c$ and increasing for $x>c$.
We know $f^{\prime}(c)=0$, so $f^{\prime}(x)>0$ for $x$ around
$c$.

Thus the first derivative test tells us $f$ does not have a local
maximum or minimum at $c$.

In general, if the first $n$-derivatives of a function $f$ equal
zero at $x=c$ and $f^{(n+1)}(c)>0$, then the graph of $f(x)$ near
$x=c$ looks like the graph of $x^{n+1}$ up close. If $f^{(n+1)}(c)<0$ then $f(x)$ looks like $-x^{n+1}$ up close. Notice the second derivative test is a particular case of this: $f^{\prime}(c)=0$ and $f^{\prime\prime}(c)>0$
implies the graph of $f(x)$ looks like the graph of $x^{2}$ up close,
so it has a minimum at $x=c$, and so on.

4.4:

There are several indeterminate forms you should learn how to deal
with.

$\frac{0}{0}$: If conditions for L'Hospital's applies, use it.

$\frac{\infty}{\infty}$: If conditions for L'Hospital's aplies, use
it.

$0\cdot\infty$: Change to $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
Choose the one that seems easier to apply L'Hospital's. Or do it both
ways as a way of checking your answer.

$\infty-\infty$: try to convert the difference into a quotient. This
might be possible by using a common denominator, rationalization (i.e.
multiplying by conjugate), or factoring out a common factor.

$0^{0}$: Using $\left[f(x)\right]^{g(x)}=e^{g(x)\ln f(x)}$ we get
$e$ raised to the indeterminate $0\cdot\infty$.

Note that one should make sure that $f(x)$ is positive, at least
for values of $x$ near $a$ (when considering the limit as $x$ goes
to $a$).

$\infty^{0}$: Using $\left[f(x)\right]^{g(x)}=e^{g(x)\ln f(x)}$
we get $e$ raised to the indeterminate $0\cdot\infty$.

$1^{\infty}$: Using $\left[f(x)\right]^{g(x)}=e^{g(x)\ln f(x)}$
we get $e$ raised to the indeterminate $\infty\cdot0$.

\#2

\#6

We can factor the numerator and cancel the trouble causing denominator
with one of the factors.

\#12

L'Hospital's

$\lim_{t\to0}f(t)=\lim_{t\to0}3e^{3t}=3$

\#14

$\lim_{\theta\to\pi/2}f(\theta)=\lim_{\theta\to\pi/2}\sin\theta(1-\sin\theta)=0(1)=0$

\#18

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{\frac{1}{\ln x}\frac{1}{x}}{1}=0$

\#42

$\lim_{x\to0^{+}}\sin x\ln x=\lim_{x\to0^{+}}\frac{\sin x}{x}\frac{\ln x}{\frac{1}{x}}=\lim\frac{\sin x}{x}\cdot\lim\frac{\ln x}{\frac{1}{x}}=1\cdot\lim\frac{1/x}{-1/x^{2}}=\lim_{x\to0^{+}}(-x)=0$

Or $\lim\sin x\ln x=\lim\frac{\ln x}{1/\sin x}=\lim\frac{1/x}{-\cos x/\sin^{2}x}=\lim\frac{\sin x}{x}\cdot\lim\frac{-\sin x}{\cos x}=1\cdot0=0$

Note: I am omiting the $x\to a$ for convenience, but you can't forget
to include this in your work.

\#52

This is $\infty-\infty$ and we factor a term to get an indeterminate
of the form $\infty\cdot0$

$\lim\left(xe^{1/x}-x\right)=\lim\left(x(e^{1/x}-1)\right)=\lim\frac{e^{1/x}-1}{1/x}=\lim\frac{e^{1/x}(-1/x^{2})}{(-1/x^{2})}=\lim e^{1/x}=1$

Note: I am omiting the $x\to a$ for convenience, but you can't forget
to include this in your work.

\#54

\#64

We use $e^{g(x)\ln(f(x))}$ and consider the limit

$\lim(2x+1)\ln\left(\frac{2x-3}{2x+5}\right)$. Remembering the importance
to identify the indeterminate form, we see this is one of the form
$\infty\cdot0$ because $\lim_{x\to\infty}\frac{2x-3}{2x+5}=1$ and
so taking the natural log we get $0$.

$\lim\frac{\ln\left(\frac{2x-3}{2x+5}\right)}{1/(2x+1)}=\lim\frac{\frac{2x+5}{2x-3}\cdot\left(\frac{2(2x+5)-2(2x-3)}{(2x+5))^{2}}\right)}{-1/(2x+1)^{2}}=-16\cdot\lim\frac{2x+5}{2x-3}\cdot\left(\frac{2x+1}{2x+5}\right)^{2}=-16\cdot1=-16$

Note: I am omiting the $x\to a$ for convenience, but you can't forget
to include this in your work

\#80

The third term poses a problem if we tried to split the limit up,
so we hope that combining the terms over a common denominator can
fix the problem

We get $\frac{\sin2x+ax^{3}+bx}{x^{3}}$. So we get indeterminate
$\frac{0}{0}$.

Then apply L'Hospital.

We get $\frac{2\cos2x+3ax^{2}+b}{3x^{2}}$

We could continue with L'Hospital if the limit of the numerator is
0. At the moment the limit is $2+b$. So assume $b=-2$.

Now apply L'Hospital.

We get $\lim\frac{-4\sin2x+6ax}{6x}=\lim\frac{-4\sin2x}{3\cdot2x}+\lim\frac{6ax}{6x}=-\frac{4}{3}\lim\frac{\sin2x}{2x}+\lim a=-\frac{4}{3}+a$

And the problem wants the limit to be $0$ so $a=\frac{4}{3}$.