Determining critical values for $y = x^3 – 12x^2 + 36x + 8$

Question

Determine the critical values for the following function, and find out the critical value that constitutes a maximum.

$$y = x^3 – 12x^2 + 36x + 8$$

Answer 1

$f\left(x\right)=x^3-12x^2+36x+8$

Find where $f^{\prime}(x)$ is equal to zero or undefined

$$f^{\prime}(x)=3x^2-24x+8=0$$ $$x=6,x=2$$ Note that Domain of $x^3-12x^2+36x+8$ is $-\infty<x<\infty$

So all the critical points are in the domain.

The critical points are $x=2,x=6$

Answer 2

Recall that to find critical points we need to consider

$$f'(x)=3x^2-24x+36=3(x-6)(x-2)=0$$

and then we have $2$ critical points at $x=2$ and $x=6$.

Then we can consider the sign for $f''(x)=6x-24$ around the critical points to find out whether they are local maximum or local minimum points, notably we have that

• $f''(2)=-12<0 \implies x=2$ is a local maximum that is $f(2)=40$

• $f''(6)=12>0 \implies x=6$ is a local minimum that is $f(6)=8$