Question

Let$f:R\to R$ and let$f\left(x\right)>0$ for all $x\in R$. Show that f(x) is strictly increasing if and only if the functionrole="math" localid="1668414567143" $g\left(x\right)=1/f\left(x\right)$ is strictly decreasing.

Short answer

f(x) is strictly increasing if and only if the function$g\left(x\right)=1/f\left(x\right)$ is strictly decreasing

Step-by-step solution

Step: 1

Suppose that f is strictly increasing. This means that $f\left(x\right)<f\left(y\right)$whenever x<y. To show that g is strictly decreasing, suppose that x<y.

Then$g\left(x\right)=1/f\left(x\right)>1/f\left(y\right)=g\left(y\right)$

Step: 2           

Conversely, suppose that g is strictly decreasing. This means that$g\left(x\right)<g\left(y\right)$ whenever x < y.

To show that f is strictly increasing, suppose that x < y.

Then$f\left(x\right)=1/g\left(x\right)>1/g\left(y\right)=f\left(y\right)$