Question

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

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