Let f : R $\to$ R be defined as f(x) = 10x + 7. Find the function g:R $\to$ R such that gof = fog = $1_{R}$ .

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Solution

It is given that f :R $\to$ R is defined as f(x) = 10x + 7.
One-one:
Let f(x) = f(y), where x, y $ϵ$ R.
$\Rightarrow$ 10x+7=10y+7
$\Rightarrow$ x = y
therefore f is a one-one function.
Onto:
For y $ϵ R$ , let y = 10x + 7
$\Rightarrow x = \frac{y - 7}{10} ϵ R$
Therefore, for any y $ϵ$ R, there exists $x = \frac{y - 7}{10} ϵ R$ such that
$f (x) = f (\frac{y - 7}{10}) = 10 (\frac{y - 7}{10}) + 7 = y - 7 + 7 = y$ .
$therefore$ f is onto.
Therefore, f is one-one and onto.
Thus, f is an invertible function.
Let us define g : R $\to$ R as g(y)= $\frac{y - 7}{10}$ .
Now, we have:
$g o f (x) = g (f (x)) = g (10 x + 7) = \frac{(10 x + 7) - 7}{10} = \frac{10 x}{10} = 10$
And,
$f o g (y) = f (g (y)) = f (\frac{y - 7}{10}) = 10 (\frac{y - 7}{10}) + 7 = y - 7 + 7 = y$
$∴ g o f = I_{R} a n d f o g = I_{R}$
Hence, the required function g : R $\to$ R is defined as $g (y) = \frac{y - 7}{10}$ .

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