Answer
f(x) is strictly increasing if and only if $\frac{1}{f(x)}$ is strictly decreasing.
Work Step by Step
A function f(x) is strictly increasing if x < y, then f(x) < f(y)
Given: $f:\cal{R} \rightarrow \cal {R}$ and f(x)>0 for all $x \in \cal {R}$
To prove: f(x) is strictly increasing if and only if $\frac{1}{f(x)}$ is strictly decreasing.
Proof:
Let us assume that f(x) is strictly increasing.
If x \frac{1}{f(y)}$$
Define $g(x)=\frac{1}{f(x)}$
Then, $$g(x)>g(y)$$.
Thus , we have derived that $g(x)=\frac{1}{f(x)}$ is strictly decreasing.
Conversely,
Let us assume that $g(x)=\frac{1}{f(x)}$ is strictly decreasing.
If x g(y)$$
Taking reciprocal we have $$\frac{1}{g(x)}
