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

##### 1 Answer

Jun 10, 2017

#### 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"#