Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 18

Given: f(x)=x^2+6x+10
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)=2x+6=0
f'(x)=2x=-6
x=-3
The critical value for the first derivative is x=-3
If f'(x)>0, the function will increase in the interval.
If f'(x)<0, the function will decrease in the interval.
Choose a value for x that is less than -3.
f'(-4)=-2 Since f'(-4)<0 the function is decreasing in the interval (-oo,-3).
Choose a value for x that is greater than 0.
f'(0)=6 Since f'(0)>0 the function is increasing in the interval (-3, oo).
Because the function changed direction from decreasing to increasing a relative minimum will occur at x=-3. The relative minimum is the point (-3, 1).