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Question
Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
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Solution
f: R → R is given by,
f(x) = 4x + 3
One-one:
Let f(x) = f(y).
⇒ 4x + 3 = 4y + 3
⇒ 4x = 4y
⇒ x = y
∴ f is a one-one function.
Onto:
For y ∈ R, let y = 4x + 3.
⇒ `x = (y - 3)/4 ∈ R`
Therefore, for any y ∈ R, there exists `x = (y - 3)/4 ∈ R` such that
`f(x) = f((y - 3)/4)`
= `4((y - 3)/4) + 3`
= y
∴ f is onto.
Thus, f is one-one and onto and therefore, f−1 exists.
Let us define g: R→ R by `g(x) = (y - 3)/4`.
Now, (gof)(x) = g(f(x))
= g(4x + 3)
= `((4x + 3) - 3)/4 `
= x
(fog)(y) = f(g(y))
= `f((y - 3)/4)`
= `4((y - 3)/4) + 3`
= y – 3 + 3
= y
∴ gof = fog = IR
Hence, f is invertible and the inverse of f is given by `f^(-1) = g(y) = (y - 3)/4`.
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