Question

If $f\left(x\right)=|x^2-4|,$ then the value of derivative of $f\left(x\right)$ at $x=-1$ is

[1 mark]

• A. $-2$
• B. $2$
• C. $-1$
• D. $1$

Answer 1

The correct option is B $2$

$f\left(x\right)=|x^2-4|$
$\Rightarrow f\left(x\right)=\{\begin{array}{cc}{x}^2-4,& x\in (-\infty ,-2]\cup [2,\infty )\\ 4-x^2,& x\in (-2,2)\end{array}$
$\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{cc}2x,& x\in (-\infty ,-2)\cup (2,\infty )\\ -2x,& x\in (-2,2)\end{array}$

$\therefore f^{\prime }(-1)=-2(-1)=2$