Chapter 3.2, Problem 31E

To determine

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

Answer to Problem 31E

The required all values of $x$ of the given function $f\left(x\right)$ is differentiable are $\left(-\infty ,-1\right)\cup \left(-1,5\right)\cup \left(5,\infty \right)$

Explanation of Solution

Given information:

The given function: $f\left(x\right)=\frac{x^3-8}{x^2-4x-5}$

Calculation:

The given function is. $f\left(x\right)=\frac{x^3-8}{x^2-4x-5}$

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

Now, find the zeros of the denominator of the given function $f\left(x\right)=\frac{x^3-8}{x^2-4x-5}$

  $\begin{array}{l}\Rightarrow x^2-4x-5=0\\ \Rightarrow \left(x+1\right)\left(x-5\right)=0\\ \Rightarrow x=-1,x=5\end{array}$

Since, the given function is rational function.

So, it is must be differentiable for all $x$ in its domain that means all real numbers except $-1$ and $5$ .

Hence, the required all values of $x$ of the given function $f\left(x\right)$ are $\left(-\infty ,-1\right)\cup \left(-1,5\right)\cup \left(5,\infty \right)$ .