Chapter 5.1, Problem 55E

a.

To determine

To graph: that function.

Explanation of Solution

Given information:

The function is $f\left(x\right)=\left\{\begin{array}{l}\sin \frac{1}{x},x>0\\ 0,x=0\end{array}\right\}$ .

Graph: to graph the function, plot some points

x	$f\left(x\right)$
$0$	$0$
$1$	$\sin 1$
$2$	$\sin \frac{1}{2}$

Now, plot these points and join them and make a smooth curve,

The graph can be observed as:

  

Interpretation: from the above graph it can be observed that $f\left(0\right)=0$ is not a local extreme value because in any open interval containing $x=0$ , there are infinitely many points where $f\text{'}\left(x\right)=1$ and where $f\text{'}\left(x\right)=-1$ .

b.

To determine

To find: the function that fails to have an extreme value at a domain endpoint.

Answer to Problem 55E

The function is $f\left(x\right)=\left\{\begin{array}{l}\left(1-x\right)\cos \frac{1}{1-x},\text{ 0}\le x<1\\ 0,x=1\end{array}\right\}$ that fails to have an extreme value at a domain endpoint.

Explanation of Solution

Given information:

The function is $f\left(x\right)=\left\{\begin{array}{l}\sin \frac{1}{x},x>0\\ 0,x=0\end{array}\right\}$ .

Calculation: the one possible function can be as on the interval $\left[0,1\right]$ ,

  $f\left(x\right)=\left\{\begin{array}{l}\left(1-x\right)\cos \frac{1}{1-x},\text{ 0}\le x<1\\ 0,x=1\end{array}\right\}$

The function has no local extreme value at $x=1$ but it is continuous on $\left[0,1\right]$ .