Question

Find intervals in which $f\left(x\right)=\frac{4x^2+1}{x}$ is increasing and decreasing.

Answer 1

Given:$f\left(x\right)=\frac{4x^2+1}{x}$

Differentiating both sides w.r.t $x$, weget

$f^{\prime }\left(x\right)=\frac{x\left(8x\right)-(4x^2+1)}{x^2}=\frac{4x^2-1}{x^2}$

$\left(a\right)$ For increasing function.

$f^{\prime }\left(x\right)\ge 0$

$\Rightarrow \frac{4x^2-1}{x^2}\ge 0$

$\Rightarrow 4x^2-1\ge 0$

$\Rightarrow (2x-1)(2x+1)\ge 0$

$\Rightarrow x\ge \frac{1}{2}$ or $x\le \frac{-1}{2}$

So, $f\left(x\right)$ is increasing on $(-\infty ,\frac{-1}{2}]\cup [\frac{1}{2},\infty )$

$\left(b\right)$For decreasing function,

$\Rightarrow \frac{4x^2-1}{x^2}\le 0$

$\Rightarrow 4x^2-1\le 0$

$\Rightarrow (2x-1)(2x+1)\le 0$

$\Rightarrow x\le \frac{1}{2}$ or $x\ge \frac{-1}{2}$

So, $f\left(x\right)$ is decreasing on $[\frac{-1}{2},\frac{1}{2}]$