Find the intervals in which the function $f\left(x\right) = \left(x+1\right)^2 \left(x-3\right)^3$ is increasing or decreasing.
$f\left(x\right) = \left(x+1\right)^2 \left(x-3\right)^3$
$\therefore$ $f'\left(x\right) = 3 \left(x+1\right)^2 \left(x-3\right)^2 + 2 \left(x+1\right) \left(x-3\right)^3$
i.e. $f'\left(x\right) = \left(x+1\right) \left(x-3\right)^2 \left[3\left(x+1\right) +2 \left(x-3\right)\right]$
i.e. $f'\left(x\right) = \left(x+1\right) \left(x-3\right)^2 \left(5x-3\right)$
Now, $\left(x-3\right)^2 > 0 \; \forall \; x$
For critical points, $f'\left(x\right) = 0$
$\implies$ $x+1 = 0$ $\implies$ $x=-1$
or $\left(x-3\right)^2 =0$ $\implies$ $x=3$
or $5x-3=0$ $\implies$ $x=\dfrac{3}{5}$
These points divide the real line into four intervals namely $\left(-\infty, -1\right)$, $\left(-1, \dfrac{3}{5}\right)$, $\left(\dfrac{3}{5},3\right)$ and $\left(3,\infty\right)$
| Interval | Sign of $\left(x+1\right)$ | Sign of $\left(5x-3\right)$ | Sign of $f'\left(x\right)$ | Nature of function f |
|---|---|---|---|---|
| $\left(-\infty,-1\right)$ | -ve | -ve | +ve | Increasing |
| $\left(-1,\dfrac{3}{5}\right)$ | +ve | -ve | -ve | Decreasing |
| $\left(\dfrac{3}{5},3\right)$ | +ve | +ve | +ve | Increasing |
| $\left(3,\infty\right)$ | +ve | +ve | +ve | Increasing |
$\therefore$ $f\left(x\right)$ is increasing in the interval $\left\{x : x<-1\right\} \cup \left\{x : x>\dfrac{3}{5}\right\} $
$f\left(x\right)$ is decreasing in the interval $\left\{x : -1 < x < \dfrac{3}{5}\right\} $