Solve this

Solution:

To find: the function $g: R \rightarrow R: g \circ f=f \circ g=I_{g}$

Formula used: (i) $g$ o $f=g(f(x))$

(ii) $f \circ g=f(g(x))$

Given: $f: R \rightarrow R: f(x)=10 x+7$

We have,

$f(x)=10 x+7$

Let $f(x)=y$

$\Rightarrow y=10 x+7$

$\Rightarrow y-7=10 x$

$\Rightarrow x=\frac{y-7}{10}$

Let $g(y)=\frac{y-7}{10}$ where $g: R \rightarrow R$

$g \circ f=g(f(x))=g(10 x+7)=\frac{(10 x+7)-7}{10}$

$=x$

$=\mathrm{I}_{\mathrm{g}}$

$f \circ g=f(g(x))=f\left(\frac{x-7}{10}\right)$

$=10\left(\frac{x-7}{10}\right)+7$

$=x-7+7$

$=x$

Clearly g o $f=f \circ g$ $=I_{g}$ Ans $) \cdot g(x)=\frac{x-7}{10}$