Question

The critical points of the function $f\left(x\right)=\frac{|x-1|}{x^2}$ are

• A. 0
• B. 1
• C. 2
• D. -1

Answer 1

Answer: (A), (B), (C)

The correct options are
A 0
B 1
C 2

for $x\ge 1$
$f\left(x\right)=\frac{x-1}{x^2}$
& $f^{\prime }\left(x\right)=\frac{-x^2+2x}{x^4}$
for $x<1$
$f\left(x\right)=\frac{-x+1}{x^2}$
& $f^{\prime }\left(x\right)=\frac{x^2-2x}{x^4}$
for critical points: $f^{\prime }\left(x\right)=0$
$\Rightarrow x=0,2$
now, $f^{\prime }(1+)=1$ & $f^{\prime }(1-)=-1$
Since, $f^{\prime }\left(x\right)$ changes sign at $x=1$
Therefore, $x=1$ is also a critical point
Thus critical points of $f\left(x\right)$ are $x=0,1,2$
Ans: A,B,C