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प्रश्न
Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
उत्तर
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 in R`
Therefore, for any y ∈ R, there exists `x= (y-3)/4 in 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 = I_R`
Hence, f is invertible and the inverse of f is given by
`f^(-1) = g(y) = (y-3)/4`
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