Question

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

Answer 1

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.

${\displaystyle \Rightarrow x=\frac{y-3}{4}\in R}$

Therefore, for any y ∈ R, there exists ${\displaystyle x=\frac{y-3}{4}\in R}$ such that

${\displaystyle f\left(x\right)=f\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)+3=y}$

∴ f is onto.

Thus, f is one-one and onto and therefore, f⁻¹ exists.

Let us define g: R→ R by ${\displaystyle g\left(x\right)=\frac{y-3}{4}}$  

Now, ${\displaystyle \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)=g(4x+3)=\frac{(4x+3)-3}{4}=x}$

${\displaystyle \left(fog\right)\left(y\right)=f\left(g\left(y\right)\right)=f\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)+3=y-3+3=y}$

${\displaystyle \therefore gof=fog=I_R}$

Hence, f is invertible and the inverse of f is given by

${\displaystyle f^{-1}=g\left(y\right)=\frac{y-3}{4}}$