Question

Find the critical points of the function f(x) = x3 - 2x2

Answer 1

Find f'(x), set it equal to zero, then solve for x

Answer 2

take the derivative, set = 0 and solve

Answer 3

hi

Answer 4

hello. so would it be 3x^2-4x=0

Answer 5

\[f'(x)=3x^2-4x=0\]Thus,\[x=4/3\]The value of y at x=4/3 is:\[f(4/3)=(4/3)^3-2(4/3)^2=-32/27\]The only critical point is (4/3,-32/27)

Answer 6

no not really

Answer 7

\[f'(x)3x^2-4x\]
\[3x^2-4x=0\]
\[x(3x-4)=0\]
\[x=0\]
\[x=\frac{4}{3}\] are the two critical numbers

Answer 8

oooops...yeah good call :)

Answer 9

Satellite he had the right said up and you said it was wrong

Answer 10

I think he was responding to my reply. I missed a critical number i.e., x=0 in my solution :(