Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 20

Given: f(x)=-3x^2-4x-2
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)=-6x-4=0
-6x=4
x=4/-6
x=-2/3
The critical value for the first derivative is x=-2/3.
If f'(x)>0, the function is increasing in the interval.
If f'(x)<0, the function is decreasing in the interval.
Choose a value for x that is less than -2/3.
f'(-1)=2 Since f'(-1)>0 the graph of the function is increasing in the interval
(-oo,-2/3).
Choose a value for x that is greater than -2/3.
f'(0)=-4 Since f'(0)<0 the graph of the function is decreasing in the interval
(-2/3, oo).
Because the direction of the function changed from increasing to decreasing a relative maximum will exist at x=-2/3. The relative maximum is the point (-2/3, -2/3).