Let $f: R \rightarrow R$ be defined as $f(x)=10x+7$. Find the function $g: R \rightarrow R$ such that $g \circ f = f \circ g = I_{R}$.

(A) Given $f(x) = 10x + 7$. For $f$ to be invertible,it must be one-one and onto.
$1$. One-one: Let $f(x_1) = f(x_2)$. Then $10x_1 + 7 = 10x_2 + 7$,which implies $10x_1 = 10x_2$,so $x_1 = x_2$. Thus,$f$ is one-one.
$2$. Onto: Let $y = 10x + 7$. Solving for $x$,we get $x = \frac{y-7}{10}$. Since for every $y \in R$,there exists $x = \frac{y-7}{10} \in R$,$f$ is onto.
Since $f$ is bijective,it is invertible. Let $g(y) = f^{-1}(y)$.
$g(y) = \frac{y-7}{10}$.
Verification:
$(g \circ f)(x) = g(f(x)) = g(10x + 7) = \frac{(10x+7)-7}{10} = \frac{10x}{10} = x = I_R(x)$.
$(f \circ g)(y) = f(g(y)) = f\left(\frac{y-7}{10}\right) = 10\left(\frac{y-7}{10}\right) + 7 = y - 7 + 7 = y = I_R(y)$.
Thus,$g(x) = \frac{x-7}{10}$.