Calculus: Early Transcendentals, Chapter 4, 4.1, Section 4.1, Problem 48

Given: f(x)=5+54x-2x^3,[0,4]
Find the critical value(s) by setting the first derivative equal to zero and solving for the x value(s).
f'(x)=54-6x^2=0
54=6x^2
9=x^2
x=3,x=-3
The critical value is x=3. The value x=-3 is not in the given interval [0, 4]. Therefore x=-3 will not be used to find the absolute maximum or absolute minimum value. Plug in the critical value x=3 and the endpoints of the interval [0, 4] into the original f(x) function.
f(0)=5
f(3)=113
f(4)=93
Examine the f(x) value to determine the absolute maximum and absolute minimum.
The absolute maximum occurs at the point (3, 113).
The absolute minimum occurs at the point (0, 5).