
William W. answered 04/10/19
Experienced Tutor and Retired Engineer
Critical numbers are where f'(x) = 0, where f'(x) DNE, and at the endpoints of the domain intervals.
Take the derivative by using the power rule:
f'(x) = 4x3 -36x
Critical Numbers where f'(x) DNE
There are none - f'(x) is continuous for all x
Critical Numbers where f'(x) = 0
4x3 -36x = 0
4x(x2 - 9) = 0
4x(x +3)(x - 3) = 0
x = 0, -3, 3 but x = -3 is not in the domain so x = 0, 3
Critical Numbers at the endpoints
x = -2
x = 7
To find global extrema values, plug in the critical numbers into the original function:
for x = -2, f(-2) = -51
for x = 0, f(0) = 5
for x = 3, f(3) = -76
for x = 7, f(7) = 1524
The global (absolute) minimum is the point (3, -76) where the absolute minimum value is -76
The global (absolute) maximum is the point (7, 1524) where the absolute maximum value is 1524