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

Question

f(x) = 2x^3 − 3x^2 − 72x + 7 , [−4, 5]

Answer 1

f'(x) = 6x^2 -6x -72

f'(0) = 6(x^2 -x-12)

0 = 6(x+ 3) (x-4)

x = -3, 4

f(-4) = 2(-4)^3 -3(-4)^2 -72(-4) +7= 139
f(-3) = 2(-3)^3 -3(-3)^2 -72(-3) +7=132
f(4) = 2(4)^3 -3(4)^2 -72(4) + 7= -145
f(5) = 2(5)^3 -3(5)^2 -72(5)+7= -178

Answer 2

f ' (x) = 6x^2 - 6x - 72
= 0 for a local max/min
x^2 - x - 12 = 0
(x-4)(x+3) = 0
x = 4, or x = -3

f(4) = 2(64) - 3(16) - 72(4) + 7 = -201
f(-3) = 2(-27) - 3(9) - 72(-3)+7 = 115

end-values:
f(-4) = 2(-64) - 2(16)( - 72(-4) + 7 = 135
f(5) = 2(125) - 3(25) - 72(5) + 7 = -178

so what do you think ?

Answer 3

looks like we both messed up in the f(x) calculations

should be

f(4) = 2(64) - 3(16) - 72(4) + 7 = -201
f(-3) = 2(-27) - 3(9) - 72(-3)+7 = 142

end-values:
f(-4) = 2(-64) - 3(16)( - 72(-4) + 7 = 119
f(5) = 2(125) - 3(25) - 72(5) + 7 = -178