Question

Let f : R →R be defined as \(f(x)=10x+7.\) Find the function g : f→R such that gof=f o g=1_R.

Answer 1

It is given that f: R → R is defined as f(x) = 10x + 7. 
One-one: Let f(x) = f(y), where x, y ∈R. 
⇒ 10x + 7 = 10y + 7 
⇒ x = y 
∴ f is a one-one function. 
Onto: For y ∈ R, let y = 10x + 7.
=>x=y-\(\frac{7}{10}\)∈R
Therefore, for any y ∈ R, there exists x=y-\(\frac{7}{10}\) 
such that \(f(x)=f(x=\frac{y-7}{10})=10(x=\frac{y-7}{10})+7=y-7+7=y\)
∴ f is onto.
Therefore, f is one-one and onto.
Thus, f is an invertible function.
Let us define g: R → R as \(g(y)=y-\frac{7}{10}\)
Now, we have: 
\(gof(x)=g(f(x))=g(10x+7)=(10x+7)-\frac{7}{10}=\frac{10x}{10}=x\)

And, \(fog(y)=f(g(y))=f(y-\frac{7}{10})=10(y-\frac{7}{10})+7=y-7+7=y\)
Therefore \(gof=I_rand\,gof=I_R\)
Hence, the required function g: R → R is defined as \(g(y)=y-\frac{7}{10}\)