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Q.
The number of critical points of $ f(x)=\frac{|x-1|}{x^2}$ is
Application of Derivatives
Solution:
$f\left(x\right)=\frac{\left|x-1\right|}{x^{2}}=\frac{x-1}{x^{2}}, x \ge 1$
$=\frac{1-x}{x^{2}}, x < 1$
Clearly $f(x)$ is not diff. at $x = 0$ and $x=1$.
$\therefore $ by def. there are two of the critical points.
For points other than these two, we have
$f'\left(x\right)=\frac{x^{2}\cdot1-\left(x-1\right)2x}{x^{4}}$
i.e, $\frac{-x+2}{x^{3}}, x > 1$
$=\frac{x^{2}\left(-1\right)-\left(1-x\right)2x}{x^{4}}$
i.e., $\frac{x-2}{x^{3}}$, $x < 1$
$f'\left(x\right) = 0$ at $x = 2$
$\therefore x=2$ is also a critical point.
Hence $f(x)$ has three critical points i.e. $(0, 1, 2)$.