Question

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

Answer 1

$$\begin{gathered} f\left(x\right)=ax+b \\ \text{Let}{x}_1,x_2\in R\text{such that}{x}_1<x_2. \\ \text{Then,} \\ {x}_1<x_2 \\ \Rightarrow a{x}_1>a{x}_2\text{(∵}\mathit{\text{a}}\text{<0)} \\ \Rightarrow a{x}_1+b>a{x}_2+b \\ \Rightarrow f\left(x_1\right)>f\left(x_2\right) \\ \text{Thus,}{x}_1<x_2 \\ \Rightarrow f\left(x_1\right)>f\left(x_2\right),\forall x_1,x_2\in R \\ \text{So,}f\left(x\right)\text{is decreasing on}R. \end{gathered}$$