f(x)=4x^3-x^4 how do you find the points of inflection, I know you have to find the second derivative but how do you figure it out using the sign charts.

Question

f(x)=4x^3-x^4  how do you find the points of inflection, I know you have to find the second derivative but how do you figure it out using the sign charts.

anonymous · Answer

if f''(x) > 0, it's concave up
f''(x) < 0, concave down

anonymous · Answer

do i have to plug in the zero's to the original funtion?

anonymous · Answer

no, you're checking the values before and after the inflection point

anonymous · Answer

can you show me how using the equation above?

anonymous · Answer

f''(x) = 24x - 12x^2 yes?

anonymous · Answer

24x - 12x^2 = 0
x = 0 and 2 yes? 
so you're checking the sign of f''(x) (-inf,0) , (0,2), and (2,inf)

anonymous · Answer

ok

anonymous · Answer

f''(x) < 0 on (-inf,0) and (2,inf) so it's concave down there.
and f''(x) > 0 on (0,2) so it's concave up there

anonymous · Answer

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anonymous · Answer

what do i do with this?

anonymous · Answer

sourwing, i dont understand what you mean on your  last messsage.

anonymous · Answer

is f''(-1) positive or negative?

anonymous · Answer

in other words, is 24x - 12x^2 positive or negative when x = -1?

anonymous · Answer

it is negative

anonymous · Answer

so it's concave down on (-inf, 0)

anonymous · Answer

do the same for the other intervals

anonymous · Answer

at 1 it is positive and at 3 it is negative right?

anonymous · Answer

yes

anonymous · Answer

ok

anonymous · Answer

I think I understand now, thankyou