Chapter 3.1, Problem 31E

To determine

To calculate:The one sided derivatives of the function $f\left(x\right)$ .

Answer to Problem 31E

The derivative of $f\left(x\right)$ is not differentiable at $x=1$ .

Explanation of Solution

Given information:

The function: $f\left(x\right)=\{\begin{array}{l}{x}^2+x,x\le 1\\ 3x-2,x>1\end{array}$

Formula used:

Definition of the derivative:

  $\frac{d}{dx}f\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Where $f\left(x\right)$ is the function and $h$ is a constant.

Calculation:

The given function be $f\left(x\right)=\{\begin{array}{l}{x}^2+x,x\le 1\\ 3x-2,x>1\end{array}$ .

Find the value of $f\left(x\right)$ at $x=1$ .

For $x\le 1$ , $f\left(x\right)=x^2+x$ .

Substitute the value 1 for $x$ .

  $\begin{array}{c}f\left(1\right)=1^2+1\\ =2\end{array}$

Now, find the right handed derivatives.

To find the value of $\frac{d}{dx}f\left(x\right)$ , use the formula $\frac{d}{dx}f\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ .

Write the above formula using the given function as:

  $\begin{array}{c}\frac{d}{dx}f\left(x\right)=\underset{h\to 1^{+}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}\\ =\underset{h\to 1^{+}}{\lim }\frac{3x-2-2}{x-1}\\ =\underset{h\to 1^{+}}{\lim }\frac{3x-4}{x-1}\\ =-\infty \end{array}$

Since, the limit does not exist. Thus, the right sided derivative of $f\left(x\right)$ is not differentiable at $x=1$ .

Hence, thederivative of $f\left(x\right)$ is not differentiable at $x=1$ .