Answer
Although f(x) is increasing on (-$\infty$, -1) and decreasing on (-$\infty$, -1), there are no critical numbers because there is a discontinuity at x = -1.
Work Step by Step
f(x) = $\frac{1}{(x+1)^{2}}$
Using the graph, it can be found that f is increasing on the interval (-$\infty$, -1) and decreasing on the interval (-$\infty$, -1)
Analytically:
f(x) = $(x+1)^{-2}$
f'(x) = (-2)($(x+1)^{-3}$
f'(x) = $\frac{-2}{(x+1)^{3}}$
Solving for critical numbers:
Set f'(x) = 0
0 = $\frac{-2}{(x+1)^{3}}$
x = Does Not Exist
There are no critical numbers. There is only a discontinuity at x = -1.
