Question

The derivative of $f\left(x\right)=3\left|2+x\right|$at the point $x_0=-3$ is

• A. $3$
• B. $-3$
• C. $0$
• D. does not exist

Answer 1

The correct option is B

$-3$

Explanation for the correct option:

Differentiating the given function:

$$\begin{gathered} f\left(x\right)=3\left|2+x\right| \\ \mathrm{We}\mathrm{know}\mathrm{that} \\ f\text{'}\left(c\right)=\underset{h\to 0}{\lim }\frac{f\left(c+h\right)-f\left(c\right)}{h} \\ (\mathrm{Differenciation}\mathrm{by}\mathrm{first}\mathrm{principle}) \\ \mathrm{Here}, \\ c=-3,So \\ f\text{'}(-3)=\underset{h\to 0}{\lim }\frac{f(-3+h)-f(-3)}{h} \\ =\underset{h\to 0}{\lim }\frac{3\left|2+\left(-3+h\right)\right|-3\left|2+\left(-3\right)\right|}{h} \\ =\underset{h\to 0}{\lim }\frac{3\left|2-3+h\right|-3\left|2-3\right|}{h} \\ f\text{'}(-3)=\underset{h\to 0}{\lim }\frac{3\left|-1+h\right|-3\left|-1\right|}{h} \\ =\underset{h\to 0}{\lim }\frac{3\left|h-1\right|-3\times 1}{h} \\ \mathrm{Since}h\to 0,soh-1<0 \\ \Rightarrow \left|h-1\right|=-\left(h-1\right) \\ =-h+1 \\ =(1-h) \\ \mathrm{Therefore} \\ {f}^{\text{'}}\left(-3\right)=\underset{h\to 0}{\lim }\frac{3\left(1-h\right)-3}{h} \\ =\underset{h\to 0}{\lim }\frac{3-3h-3}{h} \\ =\underset{h\to 0}{\lim }\frac{-3h}{h} \\ =-3 \\ Hence, \\ f\text{'}\left(-3\right)=-3 \end{gathered}$$

Therefore, the correct answer is option (B).