Question

The number of critical points of the fuction $f^{\prime }\left(x\right),$ where $f^{\prime }\left(x\right)=\frac{|x-2|}{x^3}$ are

• A. $0$
• B. $1$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

$f\left(x\right)=\{\begin{array}{c}\frac{x-2}{x^3},x>1\\ \frac{2-x}{x^3},x<1,x\ne 0\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{c}\frac{2(3-x)}{x^4},x\in (-\infty ,0)\cup (0,1)\\ \frac{2(x-3)}{x^4},x\in (1,\infty )\end{array}$

$f^{\prime \prime }\left(x\right)=\{\begin{array}{c}\frac{6(x-4)}{x^5},x\in (-\infty ,0)\cup (0,1)\\ \frac{6(4-x)}{x^5},x\in (1,3)\cup (3,\infty )\end{array}$

$f^{\prime \prime }\left(x\right)$ doesn't exits at $x=3$

Thus critical point of $f^{\prime }\left(x\right)$ is 3.