Question

Find the intervals in which the function $f\left(x\right)=\frac{x^4}{4}-x^3-5x^2+24x+12$ is
(a) strictly increasing,
(b) strictly decreasing.

Answer 1

$f\left(x\right)=\frac{x^4}{4}-x^3-5x^2+24x+12$

$f^{\prime }\left(x\right)=x^3-3x^2-10x+24$

$=(x-2)(x-4)(x+3)$

Put $f^{\prime }\left(x\right)=0$, we have

$x=-3,2,4$

So, the intervals are $(-\infty ,-3),(-3,2),(2,4)$ and $(4,\infty )$.

$\begin{array}{ccc}\text{Interval}& \text{Sign. of f(x)}& \text{Comment on}\\ & & \text{Nature of f(x)}\\ (-\infty ,-3)& (-)(-)(-)i.e.,<0& \text{f(x) is decreasing}\\ (-3,2)& (-)(-)(+)i.e.,>0& \text{f(x) is increasing}\\ (2,4)& (+)(-)(+)i.e.,<0& \text{f(x) is decreasing}\\ (4,\infty )& (+)(+)(+)i.e.,>0& \text{f(x) is increasing}\end{array}$


Hence, $f\left(x\right)$ is strictly increasing in $(-3,2)\cup (4,\infty )$,
and $f\left(x\right)$ is strictly decreasing in $(-\infty ,-3)\cup (2,4)$.