Question

Consider f: R$\to$R given by f(x) = 4x+3. Show that f is invertible. Find the inverse of f.

Answer 1

f: R $\to$ R is given by,
f(x) = 4x + 3

One-one:
Let f(x) = f(y).
$\implies$4x+3 = 4y+3
$\implies$ 4x = 4y
$\implies$x = y.
∴ f is a one-one function.

Onto:
For y ∈ R, let y = 4x + 3.
$\implies$x = $\frac {y-3}{4}$ ∈R
Therefore, for any y ∈ R, there exists x = $\frac {y-3}{4}$ ∈R such that
f(x) = f$(\frac {y-3}{4})$ = 4$(\frac {y-3}{4})$+3 = y.
∴ f is onto.
Thus, f is one-one and onto and therefore, f−1 exists.
Let us define g: R$\to$ R by g(y) = $(\frac {y-3}{4})$.
Now (g0f)(x) = g(f(x)) = g(4x+3) = $\frac {(4x+3)-3}{4}$=x.
(fog)(y) = f(g(y)) = f$(\frac {y-3}{4})$ = 4$(\frac {y-3}{4})$+3 = y-3+3 = y.
therefore gof = fog = IR

Hence, f is invertible and the inverse of f is given by
f-1(y) = g(y) = $\frac {y-3}{4}$.