Question

How do you find the critical points of `b'(x)=x^3+3x^2-4x-12`?

Answer 1

`x=-3" and " x=+-2`

Explanation:

`"find the critical points by equating " b'(x)" to zero"`

`rArrx^3+3x^2-4x-12=0`

`rArrx^2(x+3)-4(x+3)=0`

`rArr(x+3)(x^2-4)=0`

`x+3=0rArrx=-3larr" is a critical point"`

`(x-2)(x+2)=0rArrx=+-2larr" are critical points"`