Solution:

It is given that $f: \mathbf{R} \rightarrow \mathbf{R}$ is defined as $f(x)=10 x+7 .$

One-one:

Let $f(x)=f(y)$, where $x, y \in \mathbf{R}$.

$\Rightarrow 10 x+7=10 y+7 \\ $

$\Rightarrow x=y \\ $

$\therefore f$ is a one-one function.

Onto:

For $y \in \mathbf{R} \\ $

let $y=10 x+7 \\ $.

$\Rightarrow x=\frac{y-7}{10} \in \mathbf{R} \\ $ Therefore, for any $y \in \mathbf{R}$, there exists $x=\frac{y-7}{10} \in \mathbf{R}$

such that, $f(x)=f\left(\frac{y-7}{10}\right)=10\left(\frac{y-7}{10}\right)+7=y-7+7=y . \\ $

$\therefore f$ is onto.

Therefore, f is one-one and onto.

Thus, f is an invertible function.

Let us define,

$g: \mathbf{R} \rightarrow \mathbf{R} \\ $

as

$g(y)=\frac{y-7}{10} \\ $.

Now, we have:

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

And,

$\begin{aligned} &f \circ g(y)=f(g(y))=f\left(\frac{y-7}{10}\right)=10\left(\frac{y-7}{10}\right)+7=y-7+7=y \\\\ &\therefore g \circ f=\mathrm{I}_{\mathbf{R}} \text { and } f \circ g=\mathrm{I}_{\mathbf{R}} \\ \end{aligned}$

Hence, the required function $g: \mathbf{R} \rightarrow \mathbf{R} \\ $ is defined as $g(y)=\frac{y-7}{10} \\ $