Chapter 3.2, Problem 33E

To determine

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

Answer to Problem 33E

The required all values of $x$ of the given function $P\left(x\right)$ is differentiable are $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ .

Explanation of Solution

Given information:

The given function: $P\left(x\right)=\sin \left(\left|x\right|\right)-1$

Formula used:

Derivative function formula:

The differentiable function formulas to find the left hand derivative and right hand derivative are

Left hand derivative $=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ and Right hand derivative $=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ respectively.

Calculation:

The given function is. $P\left(x\right)=\sin \left(\left|x\right|\right)-1$

The given function can be written as $P\left(x\right)=\left\{\begin{array}{l}-\sin x-1,x<0\\ \sin x-1,x\ge 0\end{array}\right\}$

To find the Left hand derivative, use the differentiable function formula. $\text{Left}\text{hand}\text{derivative}=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ .

Use $P\left(x\right)$ instead of $f\left(x\right)$ .

Substitute $-\sin x-1$ for $P\left(x\right)$ and $0$ for $a$ in the above formula.

  $\begin{array}{c}\text{Left}\text{hand}\text{derivative}=\underset{x\to 0^{-}}{\lim }\frac{\left(-\sin x-1\right)-P\left(0\right)}{x-0}\\ =\underset{x\to 0^{-}}{\lim }\frac{-\sin x-1-\left(-1\right)}{x}\\ =\underset{x\to 0^{-}}{\lim }\frac{-\sin x}{x}\\ =-1\end{array}$

Again, to find the Right hand derivative, use the differentiable formula $\text{Right}\text{hand}\text{derivative}=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ .

Use $P\left(x\right)$ instead of $f\left(x\right)$ .

Substitute $\sin x-1$ for $P\left(x\right)$ and $0$ for $a$ in the above formula.

  $\begin{array}{c}\text{Right}\text{hand}\text{derivative}=\underset{x\to 0^{+}}{\lim }\frac{\left(\sin x-1\right)-P\left(0\right)}{x-0}\\ =\underset{x\to 0^{+}}{\lim }\frac{\sin x-1-\left(-1\right)}{x}\\ =\underset{x\to 0^{+}}{\lim }\frac{\sin x}{x}\\ =1\end{array}$

Since, the limits are not equal;

So, the function is not differentiable at $x=0$ .

Thus, the given function is differentiable for all real numbers except $x=0$ .

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