Find the intervals in which $f\left(x\right) = \dfrac{3}{10}x^4-\dfrac{4}{5}x^3 -3x^2 + \dfrac{36}{5}x + 11$ is increasing and in which it is decreasing.
$f\left(x\right) = \dfrac{3}{10}x^4 - \dfrac{4}{5}x^3 - 3x^2 + \dfrac{36}{5}x + 11$
$\therefore$ $f'\left(x\right) = \dfrac{6}{5}x^3-\dfrac{12}{5}x^2-6x+\dfrac{36}{5}$
i.e. $f'\left(x\right) = \dfrac{6}{5}\left(x^3 -2x^2-5x+6\right)$
i.e. $f'\left(x\right) = \dfrac{6}{5}\left(x-1\right) \left(x^2 -x-6\right)$
i.e. $f'\left(x\right) = \dfrac{6}{5}\left(x-1\right) \left(x-3\right) \left(x+2\right)$
$\therefore$ For critical points, $f'\left(x\right) = 0$
$\implies$ $\left(x-1\right) \left(x-3\right) \left(x+2\right) = 0$
$\implies$ $x = -2$, $x=1$, $x=3$
These points divide the real line into four disjoint intervals namely $\left(-\infty,-2\right)$, $\left(-2,1\right)$, $\left(1,3\right)$ and $\left(3,\infty\right)$
| Interval | Sign of $f'\left(x\right)$ | Nature of function f |
|---|---|---|
| $\left(-\infty,-2\right)$ | $< 0$ | Decreasing |
| $\left(-2,1\right)$ | $> 0$ | Increasing |
| $\left(1,3\right)$ | $ < 0$ | Decreasing |
| $\left(3,\infty\right)$ | $ > 0$ | Increasing |
$\therefore$ $f\left(x\right)$ is increasing in the intervals $\left(-2,1\right)$ and $\left(3,\infty\right)$
$f\left(x\right)$ is decreasing in the intervals $\left(-\infty,-2\right)$ and $\left(1,3\right)$