Question

Prove that function $f:R\to R,F\left(x\right)=\frac{3-2x}{7}$ in - one - one and onto. Also find $f^{-1}$

Answer 1

Given:-

$f\left(x\right)=\frac{3-2x}{7}f:R\to R$

Let $f\left(x\right)=f\left(y\right)$

$\frac{3-2x}{7}=\frac{3-2y}{7}$

$3-2x=3-2y$

$\Rightarrow x=y$

$\therefore f\left(x\right)$ is one-one.

Now,

Let $y=f\left(x\right)$

$y=\frac{3-2x}{7}$

$7y=3-2x$

$\Rightarrow x=\frac{3-7y}{2}$

$\Rightarrow g\left(x\right)=\frac{3-7x}{2}g:R\to R$

Hence $f\left(x\right)$ is one-one and onto.

$f^{-1}x=g\left(x\right)=\frac{3-7x}{2}$.