Question

Find relative extrema for f(x) = x^4

Answer 1

I'm confused.
I got f'(x) = 4x^3
So the only critical number is x = 0


Based on the graph, I expect (0,0) to be a relative min

However,
I got f''(x) = 12x^2

f''(0) = 0

I thought this means that (0,0) is neither a min nor max.

Someone explain please

Answer 2

If f''(0)>0 then it's a minimum and f''(0)<0 then it's a minimum if f'(0)=0. That is true. However don't take this to necessarily mean that f''(0)=0 implies that it's neither a max or a min. In the case of y=x^4 this just means that it's just really flat at that point, not that it's like x^3.

Kind of tricky, I know, but if the second derivative is 0 at a point, you have to check. One way is to look at points near it by checking the nearby points.

Answer 3

I see. Thanks for explaining this to me :]