Question

Let $f:R\to$ be defined as f(x) =10x +7. Find the function $g:R\to R$ such that gof =fog =$I_R$.

Answer 1

It is given that $f:R\to R$ is defined as f(x) =10x+7.
For one-one,
Let f(x) =f(y), where $x,y\in R$
$\Rightarrow 10x+7=10y+7\Rightarrow x=y$
Therefore, f is a one-one function.
For onto,
For $y\in R$, let $y=10x+7\Rightarrow x=\frac{y-7}{10}\in R$
Therefore, for any $y\in R$, therefore exists $x=\frac{y-7}{10}\in R$ such that
$f\left(x\right)=f\left(\frac{y-7}{10}\right)=10\left(\frac{y-7}{10}\right)+7=y-7+7=y$
Therefore, f is onto. Therefore, f is one-one and onto.
Thus, f is an invertible function.
Let us define g: $R\to Rasg\left(y\right)=\frac{y-7}{10}$
Now, we have gof(x)

$=g\left(f\right(x\left)\right)=g(10x+7)=\frac{(10x+7)-7}{10}=\frac{10x}{10}=xandfog\left(y\right)=f\left(g\right(y\left)\right)=f\left(\frac{y-7}{10}\right)=10\left(\frac{y-7}{10}\right)+7=y-7+7=y$
$\therefore gof=I_R$ and $fog=I_R$
Hence, the required function $g:R\to R$ is defined as $g\left(y\right)=\frac{y-7}{10}$