Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 9

Given: g(x)=x^2-2x-8
Find the critical values by setting the derivative equal to zero and solving for the x value(s).
g'(x)=2x-2=0
2x=2
x=1
The critical value is x=1.
If g'(x)>0 the function will increase over an interval.
If g'(x)<0 the function will decrease over an interval.
Choose an x value less than 1.
g'(0)=-2 Since g'(0)<0 the function will decrease in the interval (-oo, 1).
Choose an x value greater than 2.
g'(2)=2 Since g'(2)>0 the function will increase in the interval (1, oo).
Because the function changed direction from decreasing to increasing a relative minimum will exist. The relative minimum will occur at the point (1, -9).