Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 17

Given: f(x)=x^2-4x
Find the critical values of x by setting the derivative of the function equal to zero and solving for the x-value(s).
f'(x)=2x-4=0
2x=4
x=2
The critical value is at x=2.
If f'(x)>0, then the function is increasing on that interval.
If f'(x)<0, then the function is decreasing on that interval.
Choose any value for x that is less than 2.
f'(0)=2(0)-4=-4
Since f'(0)<0, the function is decreasing on the interval (-oo, 2).
Choose any value for x that is greater than 2.
f'(3)=2(3)-4=2
Since f'(3)>0, then the function is increasing on the interval (2, oo).
Since the sign of the derivative changed from negative to positive, then there will absolute minimum at x=2. The absolute minimum is the point (2, -4).