# How do you find the critical points for  f(x) = x^4 + 2x^3?

Oct 12, 2017

@ x = -(3/2), It is concave up and the critical point is low and high @ x = 0.

graph{x^4+2x^3 [-10, 10, -5, 5]}

#### Explanation:

f’(x)=4x^3+6x^2=0
$2 {x}^{2} \left(2 x + 3\right) = 0$
$x = 0 , - \left(\frac{3}{2}\right)$

f”(x)=12x^2+6(x)

f”(0)=0
f”(-(3/2))=27-9=18

@ x = -(3/2), It is concave up and the critical point is low and high @ x = 0.