The critical points of the function f'(x), where f(x) = $\frac{|x-2|}{x^3}$ are
Answer & explanation
Correct answer: option 4
$f(x)=\left\{\begin{array}{l}\frac{x-2}{x^3}, x \geq 1 \\ \frac{2-x}{x^3}, x<1, x \neq 0\end{array}\right.$
$\Rightarrow f'(x)=\left\{\begin{array}{l}\frac{2(3-x)}{x^4}, x>1 \\ \frac{2(x-3)}{x^4}, x<1, x \neq 0\end{array}\right.$
which shows that f''(x) does not exist at x = 3.
∴ Critical point of f'(x) is 3.