Calculus 1 hw please help

Question

Find the absolute maximum and absolute minimum values of f on the given interval.

f(x) = 4x³ − 12x² − 96x + 7,    [−3, 5]

Answer 1

take the derivative, set = 0 and solve for x

f'(x) = 12x^2 -24x -96 = 0

divide by 12

x^2-2x -8 = 0

factor

(x-4)(x+2) =0

x = -2 and 4

plug those numbers into the original cubic equation to get

f(-2) =143, f(4)= -313, as the local and probably absolute max and min

check the end points though

f(-3)=79 f(5)=-273

so -2 gives the absolute and local max, and 4 gives the absolute and local min