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Q.
The function $\frac{|x-1|}{x^2}$ is monotonically decreasing at the point
Application of Derivatives
Solution:
$f(x)= \begin{cases}\frac{1-x}{x^2} & x<1, x \neq 0 \\ \frac{x-1}{x^2}, & x \geq 1\end{cases}$
The given function is not differentiable at $x=1$
$f^{\prime}(x)=\begin{cases}\frac{1}{x^2}-\frac{2}{x^3}, & x < 1, & x \neq 0 \\ \frac{2}{x^3}-\frac{1}{x^2}, & x>1\end{cases}$
Now $ f^{\prime}(x) < 0 \Rightarrow \begin{cases}\frac{x-2}{x^3} < 0 & \text { given } & x < 1 \\ \frac{2-x}{x^3} < 0 & \text { when } & x>1\end{cases}$
$f(x)$ decreasing $\forall x \in(0,1) \cup(2, \infty)$ and $f(x)$ increases $\forall x \in(-\infty, 0) \cup(1,2)$
here $f(x)$ is decreasing at all points in $x \in(0,1) \cup(2, \infty)$ so will also be decreasing at $x=3$ at $x=1$ minima and at $x=2$ maxima
