Chapter 4, Problem 55RE

a.

To determine

Determine where the function is differentiable.

Answer to Problem 55RE

Differentiablefor all real numbers except x =1.

Explanation of Solution

Given information: Given function is $f(x)=\sqrt[7]{{(x-1)}^3}.$

Calculation:

f(x) is continuous for all real numbers since odd roots have a domain of all real numbers. To determine where f(x) is differentiable ,we must then determine where f’(x) is discontinuous.

  $\begin{array}{l}f(x)=\sqrt[7]{{(x-1)}^3}.\\ $ f(x)={(x-1)}^{\frac{3}{7}}.$$ f\text{'}(x)=\frac{3}{7}{(x-1)}^{-\frac{4}{7}}.$f\text{'}(x)=\frac{3}{7\sqrt[7]{{(x-1)}^4}}.\end{array}$

  $f\text{'}(x)$ is then discontinuous at x =1 since this x will cause division by 0. $f(x)$ is then differentiable for all real numbers except x =1.

b.

To determine

Determine where the function is continuous but not differentiable.

Answer to Problem 55RE

So, for x =1 $f(x)$ is continuous but not differentiable.

Explanation of Solution

Given information: Given function is $f(x)=\sqrt[7]{{(x-1)}^3}.$

Calculation:

f(x) is continuous for all real numbers since odd roots have a domain of all real numbers. To determine where f(x) is differentiable ,we must then determine where f’(x) is discontinuous.

  $\begin{array}{l}f(x)=\sqrt[7]{{(x-1)}^3}.\\ $ f(x)={(x-1)}^{\frac{3}{7}}.$$ f\text{'}(x)=\frac{3}{7}{(x-1)}^{-\frac{4}{7}}.$f\text{'}(x)=\frac{3}{7\sqrt[7]{{(x-1)}^4}}.\end{array}$

  $f\text{'}(x)$ is then discontinuous at x =1 since this x will cause division by 0. $f(x)$ is then differentiable for all real numbers except x =1.

So, for x =1 $f(x)$ is continuous but not differentiable .

c.

To determine

Determine where the function neither continuous nor differentiable.

Answer to Problem 55RE

Function neither continuous nor differentiable,nowhere.

Explanation of Solution

Given information: Given function is $f(x)=\sqrt[7]{{(x-1)}^3}.$

Calculation:

f(x) is continuous for all real numbers since odd roots have a domain of all real numbers. To determine where f(x) is differentiable ,we must then determine where f’(x) is discontinuous.

  $\begin{array}{l}f(x)=\sqrt[7]{{(x-1)}^3}.\\ $ f(x)={(x-1)}^{\frac{3}{7}}.$$ f\text{'}(x)=\frac{3}{7}{(x-1)}^{-\frac{4}{7}}.$f\text{'}(x)=\frac{3}{7\sqrt[7]{{(x-1)}^4}}.\end{array}$

  $f\text{'}(x)$ is then discontinuous at x =1 since this x will cause division by 0. $f(x)$ is then differentiable for all real numbers except x =1.

So, for x =1 $f(x)$ is continuous but not differentiable .

So, nowhere $f(x)$ neither continuous nor differentiable