Let R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that gof = fog = IR.

Let f : X → Y, where X ⊆ R, Y ⊆ R.
Consider an arbitrary element y of Y.
By definition, y = 10x + 7 for some x ∈ X.
⇒ x = \(\frac { y-7 }{ 10 }\).
Let us define g : Y → X by g(y) = \(\frac { y-7 }{ 10 }\)
= \(\frac { (10x+7)-7 }{ 10 }\) = x
Now gof(x) = g[f(x)] = \(\frac { f(x)-7 }{ 10 }\)
= \(\frac { (10x+7)-7 }{ 10 }\)
and fog(x) = f[g(x)] + 10g(y) + 7 = 10.\(\frac { y-7 }{ 10 }\) + 7 = y.
⇒ gof(x) = IR, fog(y) = IR
⇒ f is invertible and g : Y → X, such that g(y) = \(\frac { y-7 }{ 10 }\).