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Question
Let R be the set of real numbers and f: R → R be the function defined by f(x) = 4x + 5. Show that f is invertible and find f–1.
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Solution
Here the function f : R → R is defined as f (x) = 4x + 5 = y (say). Then
4x = y – 5 or x = `(y - 5)/4`.
This leads to a function g: R → R defined as
g(y) = `(y - 5)/4`.
Therefore, (gof) (x) = g(f(x) = g(4x + 5)
= `(4x + 5 - 5)/4`
= x
or
gof = IR
Similarly (fog) (y) = f(g(y))
= `f((y - 5)/4)`
= `4((y - 5)/4) + 5`
= y
or
fog = IR
Hence f is invertible and f-1 = g which is given by `f^-1 (x) = (x - 5)/4`
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