Question

All the critical points of $f\left(x\right)=\frac{|2-x|}{x^2}$ is/are:

• A. $x=0,2$
• B. $x=2,4$
• C. $x=2,-4$
• D. None of the above.

Answer 1

The correct option is D None of the above.

The critical point for $f\left(x\right)$ is a value $x_0$ where its derivative is zero. i.e., $f^{{}^{\prime }}\left(x_0\right)=0$

Given that $f\left(x\right)=\frac{|2-x|}{x^2}$

Therefore, $f^{{}^{\prime }}\left(x\right)=\frac{x^2\left(\frac{|2-x|}{2-x}\right)-|2-x|\left(2x\right)}{x^4}=0$

$⟹\frac{|2-x|(x^2-2x(2-x\left)\right)}{x^4(2-x)}=0$

$⟹|2-x|(x^2-2x(2-x\left)\right)=0$ and $x^4\ne 0$ and $2-x\ne 0$

$⟹|2-x|=0$ and $(3x-4)x=0$ and $x\ne 0$ and $x\ne 2$

$⟹x=2$ and $x=0$ and $x=\frac{4}{3}$and $x\ne 0$ and $x\ne 2$

Therefore, $x=\frac{4}{3}$ is the critical point.