Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 3

Given: g(x)=x^2-2x-8
Find the critical values for x by setting the second derivative of the function equal to zero and solving for the x value(s).
g'(x)=2x-2
g''(x)=2
The critical value for the second derivative does not exist.
A critical value will exist for the first derivative.
g'(x)=2x-2=0
2x=2
x=1
If g'(x)>0, the function is increasing in the interval.
If g'(x)<0, the function is decreasing in the interval.
Choose a value for x that is less than 1.
g'(0)=-2 Since g'(0)<0 the function is decreasing in the (-oo,1).
Choose a value for x that is greater than 1.
g'(2)=2 Since g'(2)>0 the function is increasing in the interval (1,oo).
Because the function changed direction from decreasing to increasing there exists relative minimum at x=1 and the function is concave up in the interval
(-oo,oo).