Question

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

Answer 1

@ 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`
`2x^2(2x+3)=0`
`x=0, -(3/2)`

`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.