Chapter 3.2, Problem 32E

To determine

The all value of $x$ for which the given function $h\left(x\right)$ is differentiable.

Answer to Problem 32E

The required all values of $x$ of the given function $h\left(x\right)$ is differentiable are $\left(-\infty ,2\right)\cup \left(2,\infty \right)$ .

Explanation of Solution

Given information:

The given function: $h\left(x\right)=\sqrt[3]{3x-6}+5$

Formula used:

Derivative function formula:

  $\underset{k\to 0}{\lim }\frac{h\left(x+k\right)-h\left(x\right)}{k}$

Calculation:

The given function is $h\left(x\right)=\sqrt[3]{3x-6}+5$

To find the all values of $x$ for which the function $h\left(x\right)$ is differentiable.

First, find the zeros of the given function $h\left(x\right)=\sqrt[3]{3x-6}+5$

  $\begin{array}{l}\Rightarrow 3x-6=0\\ \Rightarrow 3\left(x-2\right)=0\\ \Rightarrow x-2=0\\ \Rightarrow x=2\end{array}$

Now, check the differentiability of the given function $h\left(x\right)=\sqrt[3]{3x-6}+5$ at $x=2$ , use derivative function formula.

  $\underset{k\to 0}{\lim }\frac{h\left(x+k\right)-h\left(x\right)}{k}$

Substitute $2$ for $x$ in the above formula.

  $\begin{array}{l}\underset{k\to 0}{\lim }\frac{h\left(2+k\right)-h\left(2\right)}{k}\\ =\underset{k\to 0}{\lim }\frac{\left[\sqrt[3]{3\left(2+k\right)-6}+5\right]-5}{k}\\ =\underset{k\to 0}{\lim }\frac{\sqrt[3]{6+3k-6}+5-5}{k}\\ =\underset{k\to 0}{\lim }\frac{\sqrt[3]{3k}}{k}\\ =\sqrt[3]{3k}\underset{k\to 0}{\lim }\frac{1}{k^{2/3}}\end{array}$

Simplify the above limit further;

  $\sqrt[3]{3k}\underset{k\to 0}{\lim }\frac{1}{k^{2/3}}=\infty$

Since, the given function has a vertical tangent at $x=2$ .

So, it is must be differentiable for all $x$ in its domain that means all real numbers except $2$ .

Hence, the required all values of $x$ of the given function $h\left(x\right)$ are $\left(-\infty ,2\right)\cup \left(2,\infty \right)$ .