Question

Question: Let $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\mathit{R}$and$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{>}\mathbf{}\mathbf{0}$ let for all$\mathit{x}\mathbf{\in }\mathit{R}$ . Show that f(x) is strictly increasing if and only if the function $\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{/}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$is strictly decreasing.

Short answer

Answer:

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