Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 34

Given: f(x)=x/(x-5)
Find the critical numbers by setting the first derivative equal to zero and solving for the x value(x)
f'(x)=[(x-5)(1)-x(1)]/(x-5)^2=0
x-5-x=0
-5=0
No solution.
A critical number cannot be obtained using the first derivative. Critical values also exist where f(x) is not defined. Therefore there will be a critical number at x=5.
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 less than 5.
f'(4)=-5 Since f'(4)<0 the function is decreasing in the interval
(-oo,5).
Choose a value greater than 5.
f'(6)=-5 Since f'(6)<0 the function is decreasing in the interval (5, oo).
Because the direction of the function did not change, there are NO relative extrema.