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Question
Let f: R → R be defined as f(x) = 10x + 7. Find the function g: R → R such that g o f = f o g = 1R.
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Solution
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 -7)/10 in R `
Therefore, for any y ∈ R, there exists `x = (y-7)/10 in R`
such that
`f(x) = f((y -7)/10) = 10((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 -7)/10`
Now, we have:
`gof(x) = g(f(x)) = g(10x + 7) = ((10x + 7) - 7)/10 = 10x/10 = 10`
And
`fog(y) = f(g(y)) = f((y -7)/10) = 10 ((y-7)/10) + 7 = y - 7 + 7 = y`
Hence, the required function g: R → R is defined as `g(y) = (y - 7)/10`
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