Find the intervals in which the function $f\left(x\right) = \dfrac{4x^2 + 1}{x}$ is increasing and the intervals in which it is decreasing.
$f\left(x\right) = \dfrac{4x^2 + 1}{x}$ $\implies$ $f\left(x\right) = 4x + \dfrac{1}{x}$
$\therefore$ $f'\left(x\right) = 4 - \dfrac{1}{x^2} = \dfrac{4x^2 -1}{x^2}$
Now, $x^2 > 0 \;\; \forall \; x, \;\; x \neq 0$
$f'\left(x\right) = 0$ $\implies$ $4x^2 -1 = 0$
i.e. $x^2 = \dfrac{1}{4}$ $\implies$ $x = \dfrac{1}{2}$, $x = -\dfrac{1}{2}$
These points divide the real line into three intervals namely $\left(-\infty, -\dfrac{1}{2}\right)$, $\left(-\dfrac{1}{2}, \dfrac{1}{2}\right)$ and $\left(\dfrac{1}{2},\infty\right)$
| Interval | Sign of $\left(4x^2 - 1\right)$ | Sign of $f'\left(x\right)$ | Nature of function f |
|---|---|---|---|
| $\left(-\infty,-\dfrac{1}{2}\right)$ | +ve | +ve | Increasing |
| $\left(-\dfrac{1}{2},\dfrac{1}{2}\right)$ | -ve | -ve | Decreasing |
| $\left(\dfrac{1}{2},\infty\right)$ | +ve | +ve | Increasing |
$\therefore$ $f\left(x\right)$ is increasing in the interval $\left\{x : x < \dfrac{-1}{2}\right\} \cup \left\{x : x > \dfrac{1}{2}\right\} $
$f\left(x\right)$ is decreasing in the interval $\left\{x : \dfrac{-1}{2} < x < \dfrac{1}{2}\right\} $