Prove that f(x)=a x+b, where a, b are constants


Prove that $f(x)=a x+b$, where $a$, $b$ are constants and $a<0$ is a decreasing function on $R$.


we have,

$f(x)=a x+b, a<0$

let $\mathrm{x}_{1}, \mathrm{x}_{2} \in \mathrm{R}$ and $\mathrm{x}_{1}>\mathrm{x}_{2}$

$\Rightarrow \mathrm{ax}_{1}<\mathrm{ax}_{2}$ for some $\mathrm{a}>0$

$\Rightarrow \mathrm{ax}_{1}+\mathrm{b}<\mathrm{ax}_{2}+\mathrm{b}$ for some $\mathrm{b}$

$\Rightarrow f\left(x_{1}\right)

Hence, $x_{1}>x_{2} \Rightarrow f\left(x_{1}\right)

So, $f(x)$ is decreasing function of $R$

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