Homework 7:
3.10: 4, 14, 18, 20, 23, 28, 30, 32 a
4.1: 4, 34, 42, 53, 56, 68 b, 77
4.2: 14, 16, 18, 20, 24, 28, 32
3.10:
\#4
$f^{\prime}(x)=\frac{3}{4}x^{-1/4}$
$L(x)=f(1)+f^{\prime}(1)(x-1)=0+1(x-1)=x-1$
\#14a
Let $y=f(t)$.
Then $\frac{dy}{dt}=f^{\prime}(t)=e^{\tan\pi t}\cdot\sec^{2}(\pi t)\cdot\pi$.
Then $dy=\pi\sec^{2}(\pi t)e^{\tan\pi t}dt$
\#14b
$dy=\frac{1}{2z\sqrt{1+\ln z}}dz$
\#18a
$dy=-\sin xdx$
\#18b
$dy=-\sin\left(\frac{\pi}{3}\right)(0.05)=-\frac{\sqrt{3}}{2}(0.05)=-0.025\sqrt{3}$
\#20
$\Delta y=f(2)-f(1)=\sqrt{2}-\sqrt{1}\approx0.414$
$dy=\frac{1}{2\sqrt{x}}dx=\frac{1}{2}(1)=0.5$
\#23
We find the linearization of $f(x)=x^{5}$ at $a=2$. Then for values
$x$ near 2, we have $f(x)\approx L(x)$.
$f^{\prime}(x)=5x^{4}$, $f(2)=32$, $f^{\prime}(2)=80$
$L(x)=32+80(x-2)$
$(2.001)^{5}\approx L(2.001)=32+80(2.001-2)=32+80(.001)=32+.08=32.08$
Note that computing $(2.001)^{5}$ on the calculator gives approximately
$32.08008$.
\#28
We find the linearization of $f(x)=\sqrt{x}$ at $a=100$.
$f^{\prime}(x)=\frac{1}{2}x^{-1/2}$, $f(100)=10$, $f^{\prime}(100)=\frac{1}{20}$
$L(x)=10+\frac{1}{20}(x-100)$
$\sqrt{99.8}\approx L(99.8)=10+\frac{1}{20}(99.8-100)=10+\frac{1}{20}(-0.2)=10+(-0.01)=9.99$.
Note that computing $\sqrt{99.8}$ on the calculator gives approximately
$9.98999499$.
\#30
$f^{\prime}(x)=6x^{5}$
$L(x)=1+6(x-1)$
$L(1.01)=1.06$
$1.01$ is reasonably close to $1$ and so the approximation is reasonable.
\#32a
The linearizations of all three functions is the same $L(x)=1-2x$.
This is because they have the same function values and derivative
values at $a=0$.
4.1:
\#4
a: absolute minimum
b: neither
c: neither
d: local minimum
r: absolute maximum
s: local minimum
\#34
Note: In class I had the incorrect notion of critical numbers. See
page 274 for the book's definition.
The derivative does not exist at $t=\frac{4}{3}$
There are no points at which the derivative equals 0.
Thus the set of critical numbers is $\{\frac{4}{3}\}$
\#42
Note: In class I had the incorrect notion of critical numbers. See
page 274 for the book's definition.
$f(\theta)=4\theta-\tan\theta$ and it's $f^{\prime}(\theta)=4-\sec^{2}\theta$
exists everywhere on the domain of $f$.
We are left to find points at which the derivative equals 0.
$f^{\prime}(\theta)=0$ implies $\sec^{2}\theta=4$ implies $\sec\theta=\pm2$
implies $\cos\theta=\pm\frac{1}{2}$ implies $\theta=\frac{\pi}{3}+\pi k$
or $\frac{2\pi}{3}+\pi k$ for $k\in\mathbb{Z}$.
One could also write $\theta=\frac{\pi}{3}+2\pi k$, $\frac{2\pi}{3}+2\pi k$,
$\frac{4\pi}{3}+2\pi k$, $\frac{5\pi}{3}+2\pi k$ for $k\in\mathbb{Z}$.
\#53
Use the closed interval method on page 275.
\#56
Use the closed interval method on page 275.
\#68b
$f^{\prime}(x)$ exists on the domain of $f$ which is $-2\leq x\leq0$
and $f^{\prime}(x)=1+2\sin x$.
So we find numbers where $f^{\prime}(x)=0$. Thus we solve $\sin x=-\frac{1}{2}$.
In general we would have $x=-\frac{\pi}{6}+2\pi k$ , $-\frac{5\pi}{6}+2\pi k$
for $k\in\mathbb{Z}$, but our domain is $-2\leq x\leq0$.
Thus we only have $x=-\frac{\pi}{6}$ and $x=-\frac{5\pi}{6}$.
The values of $f$ at these points are $-\frac{\pi}{6}-2\frac{\sqrt{3}}{2}=-\frac{\pi}{6}-\sqrt{3}$
and $-\frac{5\pi}{6}-2\left(-\frac{\sqrt{3}}{2}\right)=-\frac{5\pi}{6}+2\sqrt{3}$
Now we compute $f$ at the endpoints. $f(0)=0-2(1)=-2$. $f(-2)=-2-2\cos(-2)$
Use a calculator to see which value is the absolute maximum and which
is the absolute minimum.
Note, had \#68a been assigned, one could see from the graph that the
values we'd want to compute is the maximum at $x=-2$ and the minimum
at $x=-\frac{\pi}{6}$.
\#77
Read the proof on page 273. The end of the proof tells you how to
prove Fermat's Theorem for the case of a local minimum.
4.2:
\#14
$f$ is continuous on the closed interval {[}1,4{]}
$f^{\prime}$ exists on the open interval $(1,4)$.
Thus the conditions hold.
$f^{\prime}(x)=\frac{(x+2)-x}{(x+2)^{2}}=\frac{2}{(x+2)^{2}}$
Then $f^{\prime}(c)=\frac{f(4)-f(1)}{4-1}=\frac{\frac{2}{3}-\frac{1}{3}}{3}=\frac{1}{9}$
Then $(x+2)^{2}=18$
$(x+2)=\pm3\sqrt{2}$
$x=-2\pm3\sqrt{2}$.
But only $x=-2+3\sqrt{2}\in(1,4)$.
\#16
It doesn't contradict the theorem because $f^{\prime}(x)$ does not
exist at $x=\frac{1}{2}\in(0,3)$.
\#18
One way to do this problem is to suppose it had more than one. Then
$f(x_{1})=f(x_{2})=0$ for two distinct numbers $x_{1}$, $x_{2}\in\mathbb{R}$.
Let's say $x_{1}<x_{2}$.
Because $f(x)=2x-1-\sin x$ is differentiable on $\mathbb{R},$ we
use the mean value theorem to imply that there exists a $c\in(x_{1},x_{2})$
such that $f^{\prime}(c)=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}=0$.
But $f^{\prime}(x)=2-\cos x>0$. (There are various ways to argue
and/or see that $2-\cos x>0$, let me know if you don't know one way
to reason this)
Thus, we've reached a contradiction, implying that there is at most
one zero. Use the intermediate value theorem to obtain that $f$ has
at least one zero.
\#20
Approach the problem like the solution to \#18. Suppose there were
at least three. Then suppose $x_{1}
only has one zero. Contradiction. We conclude there are at most two
zeros.
In general, the same argument shows that for $f$ differentiable on
$\mathbb{R}$, if $f^{\prime}$has $n$ distinct real roots, then
$f$ has at most $n+1$ distinct real roots.
\#24
One can imitate Example 5 on page 284, or prove the exercise by contradiction
which I showed in section.
\#28
Apply the mean value theorem on the interval $[-b,b]$.
\#32
As the exercise indicates, follow example 6.
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