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Problem 93 Hard Difficulty

The three cases in the First Derivative Test cover the situations one commonly encounters but do not exhaust all possibilities. Consider the functions $ f $, $ g $, and $ h $ whos values at $ 0 $ are all $ 0 $ and, for $ x \not= 0 $,
$$ f(x) = x^4 \sin \frac{1}{x} $$ $$ g(x) = x^4 \left(2 + \sin \frac{1}{x} \right) $$
$$ h(x) = x^4 \left(-2 + \sin \frac{1}{x} \right) $$
(a) Show that $ 0 $ is a critical number of all three functions but their derivatives change sign infinitely often on both sides of $ 0 $.
(b) Show that $ f $ has neither a local maximum nor a local minimum at $ 0 $, $ g $ has a local minimum, and $ h $ has a local maximum.

Answer

(a) $f(x)=x^{4} \sin \frac{1}{x} \Rightarrow f^{\prime}(x)=x^{4} \cos \frac{1}{x}\left(-\frac{1}{x^{2}}\right)+\sin \frac{1}{x}\left(4 x^{3}\right)=4 x^{3} \sin \frac{1}{x}-x^{2} \cos \frac{1}{x}$
\[
g(x)=x^{4}\left(2+\sin \frac{1}{x}\right)=2 x^{4}+f(x) \Rightarrow g^{\prime}(x)=8 x^{3}+f^{\prime}(x)
\]
$h(x)=x^{4}\left(-2+\sin \frac{1}{x}\right)=-2 x^{4}+f(x) \Rightarrow h^{\prime}(x)=-8 x^{3}+f^{\prime}(x)$
It is given that $f(0)=0,$ so $f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0} \frac{x^{4} \sin \frac{1}{x}-0}{x}=\lim _{x \rightarrow 0} x^{3} \sin \frac{1}{x} .$ since
$-\left|x^{3}\right| \leq x^{3} \sin \frac{1}{x} \leq\left|x^{3}\right|$ and $\lim _{x \rightarrow 0}\left|x^{3}\right|=0,$ we see that $f^{\prime}(0)=0$ by the Squeeze Theorem. Also, $g^{\prime}(0)=8(0)^{3}+f^{\prime}(0)=0$ and $h^{\prime}(0)=-8(0)^{3}+f^{\prime}(0)=0,$ so 0 is a critical number of $f, g,$ and $h$
For $x_{2 n}=\frac{1}{2 n \pi}[n \text { a nonzero integer }], \sin \frac{1}{x_{2 n}}=\sin 2 n \pi=0$ and $\cos \frac{1}{x_{2 n}}=\cos 2 n \pi=1,$ so $f^{\prime}\left(x_{2 n}\right)=-x_{2 n}^{2}<0$
\[
\text { For } x_{2 n+1}=\frac{1}{(2 n+1) \pi}, \sin \frac{1}{x_{2 n+1}}=\sin (2 n+1) \pi=0 \text { and } \cos \frac{1}{x_{2 n+1}}=\cos (2 n+1) \pi=-1, \text { so }
\]
$f^{\prime}\left(x_{2 n+1}\right)=x_{2 n+1}^{2}>0 .$ Thus, $f^{\prime}$ changes sign infinitely often on both sides of 0 .
Next, $g^{\prime}\left(x_{2 n}\right)=8 x_{2 n}^{3}+f^{\prime}\left(x_{2 n}\right)=8 x_{2 n}^{3}-x_{2 n}^{2}=x_{2 n}^{2}\left(8 x_{2 n}-1\right)<0$ for $x_{2 n}<\frac{1}{8},$ but
$g^{\prime}\left(x_{2 n+1}\right)=8 x_{2 n+1}^{3}+x_{2 n+1}^{2}=x_{2 n+1}^{2}\left(8 x_{2 n+1}+1\right)>0$ for $x_{2 n+1}>-\frac{1}{8},$ so $g^{\prime}$ changes sign infinitely often on both
sides of 0
\[
\text { Last, } h^{\prime}\left(x_{2 n}\right)=-8 x_{2 n}^{3}+f^{\prime}\left(x_{2 n}\right)=-8 x_{2 n}^{3}-x_{2 n}^{2}=-x_{2 n}^{2}\left(8 x_{2 n}+1\right)<0 \text { for } x_{2 n}>-\frac{1}{8} \text { and }
\]
$h^{\prime}\left(x_{2 n+1}\right)=-8 x_{2 n+1}^{3}+x_{2 n+1}^{2}=x_{2 n+1}^{2}\left(-8 x_{2 n+1}+1\right)>0$ for $x_{2 n+1}<\frac{1}{8},$ so $h^{\prime}$ changes sign infinitely often on both
sides of 0
(b) $f(0)=0$ and since $\sin \frac{1}{x}$ and hence $x^{4} \sin \frac{1}{x}$ is both positive and negative inifinitely often on both sides of $0,$ and
arbitrarily close to $0, f$ has neither a local maximum nor a local minimum at 0
since $2+\sin \frac{1}{x} \geq 1, g(x)=x^{4}\left(2+\sin \frac{1}{x}\right)>0$ for $x \neq 0,$ so $g(0)=0$ is a local minimum.
since $-2+\sin \frac{1}{x} \leq-1, h(x)=x^{4}\left(-2+\sin \frac{1}{x}\right)<0$ for $x \neq 0,$ so $h(0)=0$ is a local maximum.

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