SOLUTION: Let f(x) an always positive function such that du/dx < 0 for all real numbers. A) Let h(x) = [f(x)]^2. For what values of x will h(x) be increasing. B) Let j(x) = f(f(x)). For w

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Let f(x) an always positive function such that du/dx < 0 for all real numbers. A) Let h(x) = [f(x)]^2. For what values of x will h(x) be increasing. B) Let j(x) = f(f(x)). For w      Log On


   

Question 1027568: Let f(x) an always positive function such that du/dx < 0 for all real numbers.
A) Let h(x) = [f(x)]^2. For what values of x will h(x) be increasing.
B) Let j(x) = f(f(x)). For what values of x will j(x) be increasing.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Let df%2Fdx+%3C+0
A) h%28x%29+=+%28f%28x%29%29%5E2 ==> h'(x) = 2f(x)f'(x) < 0, because f(x) > 0 always and f'(x) < for all real numbers x.
Thus h(x) is decreasing for all real numbers x. Nowhere will it be increasing.

B) j(x) = f(f(x)) = (fof)(x) ==> j'(x) = %28df%28f%28x%29%29%2Fdf%28x%29%29%2A%28df%28x%29%2Fdx%29.
Now the domain of fof is a subset of the domain of f(x), hence by hypothesis, df%28f%28x%29%29%2Fdf%28x%29+%3C+0. Also df%28x%29%2Fdx%3C0 again by hypothesis.
==> j'(x) > 0, and so j(x) will always be increasing in the domain of (fof)(x).