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 = I_R 

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