Question

determine the critical points of f(x)=8x³+81x²-42x-8

Answer 1

$$f(x)=8x^3+81x^2-42x-8$$

Domain: $(-\infin, \infin)$

Find the first derivative

$$f'(x)=24x^2+162x-42$$

Find the critical number(s)

$$f'(x)=0=>24x^2+162x-42=0$$

$$4x^2+27x-7=0$$

$$D=(27)^2-4(4)(-7)=841$$

$$x=\dfrac{-27\pm\sqrt{841}}{2(4)}$$

$$x_1=\dfrac{-27-29}{8}=-7$$

$$x_2=\dfrac{-27+29}{8}=\dfrac{1}{4}$$

Critical numbers: $-7, \dfrac{1}{4}.$

If $x<-7, f'(x)>0, f(x)$ increases.

If $-7<x<\dfrac{1}{4}, f'(x)<0, f(x)$ decreases.

If $x>\dfrac{1}{4}, f'(x)>0, f(x)$ increases.

$$f(-7)=8(-7)^3+81(-7)^2-42(-7)-8=1511$$

$$f(\dfrac{1}{4})=8(\dfrac{1}{4})^3+81(\dfrac{1}{4})^2-42(\dfrac{1}{4})-8=-\dfrac{213}{16}$$

Local maximum: $(-7, 1511)$

Local minimum: $(\dfrac{1}{4}, -\dfrac{213}{16})$