Question

Find the interval of monotonocity of the function $f\left(x\right)=\frac{\left|x-1\right|}{x^2}.$

Answer 1

The interva; of monotonicity for the function $f\left(x\right)=\frac{|x-1|}{x^2}$

A function & defined on an interval is increasing on (a, b) if for $x_1,x_2\in (a,b)x_1\le x_2$ which implies $f\left(x_1\right)\le f\left(x_2\right)$ ; and it is considered to decreasing on $(a,b)$ if for $x_1,x_2\in (a,b)x_1\ge x_2$

which implies $f\left(x_1\right)\ge f\left(x_2\right)$.

Lets find the critical points where the function is define and whether its derivative is zero or not defined

$f^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{x-1}{x^2}\right)$

Applying the quotient rule - $\frac{u}{v}=\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$

$f^{\prime }\left(x\right)=\frac{\frac{d}{dx}(x-1)x^2-\frac{d}{dx}{x}^2(x-1)}{(x^2{)}^2}$

$\Rightarrow \frac{(1-0)x^2-\left[2x\right(x-1\left)\right]}{(x^2{)}^2}$

$\Rightarrow \frac{x^2-2x^2+2x}{(x^2{)}^2}$

$\Rightarrow \frac{-x^2+2x}{x^4}$

$\Rightarrow \frac{-x^2+2x}{x.x^3}$

$\Rightarrow \frac{-x+2}{x^3}$

Solving for $\frac{-x+2}{x^3}=0$

$-x+2=0$

$\Rightarrow x=2$

For finding undefined points of $f^{\prime }\left(x\right)$, we take the denominator of $f^{\prime }\left(x\right)$ and compare it to zero.

$f^{\prime }\left(x\right)=\frac{-x+2}{x^3}$

$x^3=0$

$\therefore x=0$

Now, identify critical points which are not in the f(x) demain.

$f\left(x\right)=\frac{x-1}{x^2}$

Comparing the denominator to zero, we get the undefined points of $f\left(x\right)$ , which are

$x^2=0$

$\therefore x=0$

$\therefore$ The function domain is $x<0$ or $x>0$

And the function $f\left(x\right)=\frac{x-1}{x^2}$ is not defined at $x=0$, therefore we get $x=2$

$\therefore$ The intervals of monotonicity are

$-\infty <x<0$

$0<x<2$ and

$2<x<\infty$