Chapter 3.2, Problem 36E

To determine

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

Answer to Problem 36E

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

Explanation of Solution

Given information:

The given function: $C\left(x\right)=x\left|x\right|$ .

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. $C\left(x\right)=x\left|x\right|$ .

The given function can be written as

  $\begin{array}{c}C\left(x\right)=\{\begin{array}{l}x\left(-x\right),x<0\\ x\left(x\right),x\ge 0\end{array}\\ =\{\begin{array}{c}-x^2,x<0\\ x^2,x\ge 0\end{array}\end{array}$

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 $C\left(x\right)$ instead of $f\left(x\right)$ .

Substitute $-x^2$ for $C\left(x\right)$ and $0$ for $a$ in the above formula.

  $\begin{array}{c}\text{Left}\text{hand}\text{derivative}=\underset{x\to 0^{-}}{\lim }\frac{-x^2-C\left(0\right)}{x-0}\\ =\underset{x\to 0^{-}}{\lim }\frac{-x^2-0}{x}\\ =\underset{x\to 0^{-}}{\lim }\frac{-x^2}{x}\\ =\underset{x\to 0^{-}}{\lim }\left(-x\right)\end{array}$

Simplify, the above limit.

  $\underset{x\to 0^{-}}{\lim }\left(-x\right)=0$

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 $C\left(x\right)$ instead of $f\left(x\right)$ .

Substitute $x^2$ for $C\left(x\right)$ and $0$ for $a$ in the above formula.

  $\begin{array}{c}\text{Right}\text{hand}\text{derivative}=\underset{x\to 0^{+}}{\lim }\frac{x^2-C\left(0\right)}{x-0}\\ =\underset{x\to 0^{+}}{\lim }\frac{x^2-0}{x}\\ =\underset{x\to 0^{+}}{\lim }\frac{x^2}{x}\\ =\underset{x\to 0^{+}}{\lim }x\end{array}$

Simplify, the above limit.

  $\underset{x\to 0^{+}}{\lim }x=0$

Since, the limits are equal;

So, the function is differentiable at $x=0$ .

Thus, the given function is differentiable for all real numbers.

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