Question

Show that the function given by $f\left(x\right)=7x-3$ is strictly increasing on $R$.

Answer 1

Let $x_1$ and $x_2$ be two real numbers such that $x_1<x_2$
Multiplying both sides by 7, we get
$7x_1<7x_2$
Subtracting both sides by $3$, we get
$7x_1-3<7x_2-3$
$\Rightarrow f\left(x_1\right)<f\left(x_2\right)$
Thus whwn $x_1<x_2$, we get $f\left(x_1\right)<f\left(x_2\right)$
So, $f\left(x\right)$ is strictly increasing on $R$

Alternate :
Given : $f\left(x\right)=7x-3$
and $f^{\prime }\left(x\right)=7>0,\forall x\in R$
We know that for any strictly incrasing function $f\left(x\right),$ $f^{\prime }\left(x\right)>0$ for all $x\in R,$