Question

If f is a function increasing for all values of x, g is a function decreasing for all values of x, describe the behavior of h(x) = f(g(-x)).?

a. always increasing
b. always decreasing
c. increasing if x>0, decreasing if x<0
d. decreasing if x>0, increasing if x<0

Answer 1

Answer:

$h(x)$ is increasing.

Step-by-step explanation:

I'll try to give you an intuition - although informal, of course. If $g(x)$ is a decreasing function, then $g(-x)$ means to read the graph of $g(x)$ "backwards", and thus making it increasing. At that point, $f(g(-x))$ is a composition of two increasing function, and is thus increasing.

Now, for a formal answer, let's simply pick two inputs $x_1<x_2$ (we will assume that the range of $g(x)$ is part of the domain of $f(x)$, otherwise you couldn't compute $f(g(-x))$)

Since $g(x)$ is decreasing, we have

$x_1<x_2 \implies g(x_1)>g(x_2)$

But we want to compute $g(-x)$, so we have

$x_1<x_2 \implies -x_1>-x_2 \implies g(-x_1)<g(-x_2)$

Since $f(x)$ is increasing, we have

$x_1<x_2 \implies f(x_1)<f(x_2)$

and thus

$g(-x_1)<g(-x_2) \implies f(g(-x_1))<f(g(-x_2))$

So, we've shown that, given two inputs $x_1<x_2$, the images are in the same order: $f(g(-x_1))<f(g(-x_2))$

This, by definition, means that $f(g(-x))$ is an increasing function.