Chapter 3.1, Problem 41E

To determine

The correct value of the right hand derivative of $f$ at $x=0$ from (A) $-2$ (B) $0$ (C) $2$ (D) $\infty$ (E) $-\infty$ .

Answer to Problem 41E

The correct answer is (C)

Explanation of Solution

Given information:

The function: $f\left(x\right)=\{\begin{array}{l}{x}^2-1,x<0\\ 2x-1,x\ge 0\end{array}$

Value of $x=0$

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:

For right hand derivative $f\left(x\right)=2x-1,x\ge 0$ .

For the right sided limit, the limit of $f\left(0+h\right)$ derived for $x\ge 0$ . But for $f\left(0\right)$ , use the derivative law,

Now, find the derivative

Differentiate each side with respect to $x$ , use the derivative law, $\frac{d}{dx}f\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ .

  $\begin{array}{c}\frac{d}{dx}f\left(0\right)=\underset{h\to 0^{-}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{2h-1-\left(-1\right)}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{2h}{h}\\ =2\end{array}$

Thus, $f\text{'}\left(0\right)=2$ for $x=0$ .

Hence, the correct answer is (C)