Chapter 3.2, Problem 35E

To determine

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

Answer to Problem 35E

The required all values of $x$ of the given function $g\left(x\right)$ is differentiable are $\left(-3,0\right)\cup \left(3,\infty \right)$ .

Explanation of Solution

Given information:

The given function: $g\left(x\right)=\{\begin{array}{l}{\left(x+1\right)}^2,x\le 0\\ 2x+1,0<x<3\\ {\left(4-x\right)}^2,x\ge 3\end{array}$

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. $g\left(x\right)=\{\begin{array}{l}{\left(x+1\right)}^2,x\le 0\\ 2x+1,0<x<3\\ {\left(4-x\right)}^2,x\ge 3\end{array}$

Since, the function is defined in terms of polynomial.

So, it is differentiable everywhere except possibly at $x=0$ and at $x=3$ .

Check at $x=0$ .

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

Substitute ${\left(x+1\right)}^2$ for $g\left(x\right)$ and $0$ for $a$ in the above formula.

  $\begin{array}{c}\text{Left}\text{hand}\text{derivative}=\underset{x\to 0^{-}}{\lim }\frac{{\left(x+1\right)}^2-g\left(0\right)}{x-0}\\ =\underset{x\to 0^{-}}{\lim }\frac{x^2+2x+1-1}{x}\\ =\underset{x\to 0^{-}}{\lim }\frac{x^2+2x}{x}\\ =\underset{x\to 0^{-}}{\lim }\left(x+2\right)\end{array}$

Simplify, the above limit.

  $\underset{x\to 0^{-}}{\lim }\left(x+2\right)=2$

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

Substitute $2x+1$ for $g\left(x\right)$ and $0$ for $a$ in the above formula.

  $\begin{array}{c}\text{Right}\text{hand}\text{derivative}=\underset{x\to 0^{+}}{\lim }\frac{\left(2x+1\right)-g\left(0\right)}{x-0}\\ =\underset{x\to 0^{+}}{\lim }\frac{2x+1-1}{x}\\ =\underset{x\to 0^{+}}{\lim }\frac{2x}{x}\\ =2\end{array}$

Since, the limits are equal;

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

Again, check at $x=3$ .

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

Substitute $2x+1$ for $g\left(x\right)$ and $3$ for $a$ in the above formula.

  $\begin{array}{c}\text{Left}\text{hand}\text{derivative}=\underset{x\to 3^{-}}{\lim }\frac{\left(2x+1\right)-g\left(3\right)}{x-3}\\ =\underset{x\to 3^{-}}{\lim }\frac{2x+1-7}{x-3}\\ =\underset{x\to 3^{-}}{\lim }\frac{2\left(x-3\right)}{x-3}\\ =2\end{array}$

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

Substitute ${\left(4-x\right)}^2$ for $Q\left(x\right)$ and $3$ for $a$ in the above formula.

  $\begin{array}{c}\text{Right}\text{hand}\text{derivative}=\underset{x\to 3^{+}}{\lim }\frac{{\left(4-x\right)}^2-g\left(3\right)}{x-3}\\ =\underset{x\to 3^{+}}{\lim }\frac{{\left(4-x\right)}^2-1}{x-3}\\ =\underset{x\to 3^{+}}{\lim }\frac{\left(4-x-1\right)\left(4-x+1\right)}{x-3}\\ =\underset{x\to 3^{+}}{\lim }\frac{-\left(x-3\right)\left(5-x\right)}{x-3}\end{array}$

Simplify the limit function further;

  $\begin{array}{c}\underset{x\to 3^{+}}{\lim }\frac{-\left(x-3\right)\left(5-x\right)}{x-3}=\underset{x\to 3^{+}}{\lim }-\left(5-x\right)\\ =-\left(5-3\right)\\ =-2\end{array}$

Since, the limits are not equal;

So, the function is not differentiable at $x=3$ .

Thus, the given function is differentiable for all real numbers except $x=3$ .

Hence, the required all values of $x$ of the given function $g\left(x\right)$ is differentiable are $\left(-\infty ,3\right)\cup \left(3,\infty \right)$ .