Question

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

Answer 1

We have

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

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

As x = 2 satisfies the above equation. Therefore, (x − 2) is a factor.

On performing long division

${\displaystyle \frac{x^3-3x^2-10x+24}{x-2}=x^2-x-12=(x+3)(x+4)}$ 

${\displaystyle \Rightarrow f\prime \left(x\right)=(x-2)(x+3)(x-4)}$

Here, the critical points are 2, −3, and 4.
The possible intervals are (−∞, −3), (−3, 2), (2, 4), (4, ∞)

a) For f(x) to be strictly increasing, we must have

f'(x) > 0

${\displaystyle \Rightarrow (x-2)(x+3)(x-4)>0}$

${\displaystyle \Rightarrow x\in (-3,2)U(4,\infty )}$

So, f(x) is strictly increasing on x∈(−3, 2) ∪ (4, ∞).

b) For f(x) to be strictly decreasing, we must have 

f'(x) < 0

${\displaystyle \Rightarrow (x-2)(x+3)(x-4)<0}$

${\displaystyle \Rightarrow x\in (-\infty ,-3)U(2,4)}$`

So, f(x) is strictly decreasing on x∈(−∞, −3) ∪ (2, 4).