Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 19

Given: f'(x)=-2x^2+4x+3
Find the critical values for x by setting the first derivative of the function equal to zero and solving for the x value(s).
f'(x)=-4x+4=0
-4x=-4
x=1
The critical value for the first derivative is x=1.
If f'(x)>0, the function increases in the interval.
If f'(x)<0, the function decreases in the interval.
Choose a value for x that is less than 1.
f'(0)=4 Since f'(0)>0 the function increases in the interval (-oo, 1).
Choose a value for x that is greater than 1.
f'(2)=-4 Since f'(2)<0 the function decreases in the interval (1, oo). Because the function changed directions from increasing to decreasing a relative maximum will exist. The relative maximum is the point (1, 5).

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