Question

Locate the absolute extrema of f(x)=x^3 - (9/2)x^2 on the interval [-1,4]

Answer 1

absolute max at x = -1 and min at x = 3 ??????

Answer 2

take the derivative, set it equal to zero and solve to find the possibilities. then check the endpoints as well

Answer 3

\[f'(x)=3x^2-9x\]
\[3x^2-9x=0\]
\[3(x^2-3)=0\]
\[x=\pm\sqrt{3}\] unless i made a mistake

Answer 4

so only candiate in your interval is \(\sqrt{3}\) since \(-\sqrt{3}<-1\)

Answer 5

*candidate, aka critical point

Answer 6

ok wait

Answer 7

3x^2 - 9x =0 is 3x(x - 3) =0 ? so x=0, x=3

Answer 8

right? or what did i do wrong?

Answer 9

yikes i must be tired.

Answer 10

its ok lol

Answer 11

yes the critical point is at \(x=3\) so your last job is to compute \(f(-1), f(3), f(4)\)
biggest is max, smallest is min