Question

If f : Q → Q, g : Q → Q are two functions defined by f(x) = 2 x and g(x) = x + 2, show that f and g are bijective maps. Verify that (gof)⁻¹ = f⁻¹ og ⁻¹.

Answer 1

Injectivity of f:
Let x and y be two elements of domain (Q), such that
f(x) = f(y)
$\Rightarrow$2x = 2y
$\Rightarrow$x = y
So, f is one-one.

Surjectivity of f:
Let y be in the co-domain (Q), such that f(x) = y.

$$\begin{gathered} \Rightarrow 2x=y \\ \Rightarrow x=\frac{y}{2}\in Q\left(\text{domain}\right) \end{gathered}$$

$\Rightarrow$ f is onto.
So, f is a bijection and, hence, it is invertible.

Finding f ^-1:
$$\begin{gathered} \text{Let}{f}^{-1}\left(x\right)=y...\left(1\right) \\ \Rightarrow x=f\left(y\right) \\ \Rightarrow x=2y \\ \Rightarrow y=\frac{x}{2} \\ \text{So,}{f}^{-1}\left(x\right)=\frac{x}{2}\left(\text{from}\left(1\right)\right) \end{gathered}$$

Injectivity of g:
Let x and y be two elements of domain (Q), such that
g(x) = g(y)
$\Rightarrow$x + 2 = y + 2
$\Rightarrow$x = y
So, g is one-one.

Surjectivity of g:
Let y be in the co domain (Q), such that g(x) = y.

$$\begin{gathered} \Rightarrow x+2=y \\ \Rightarrow x=2-y\in Q\left(\text{domain}\right) \end{gathered}$$

$\Rightarrow$ g is onto.
So, g is a bijection and, hence, it is invertible.

Finding g ^-1:
$$\begin{gathered} \text{Let}{g}^{-1}\left(x\right)=y...\left(2\right) \\ \Rightarrow x=g\left(y\right) \\ \Rightarrow x=y+2 \\ \Rightarrow y=x-2 \\ \text{So,}{g}^{-1}\left(x\right)=x-2\left(\text{From}\left(2\right)\right) \end{gathered}$$

Verification of (gof)⁻¹ = f⁻¹ og ⁻¹:

$$\begin{gathered} f\left(x\right)=2x;g\left(x\right)=x+2 \\ \mathrm{and}{f}^{-1}\left(x\right)=\frac{x}{2};g^{-1}\left(x\right)=x-2 \\ \mathrm{Now},\left(f^{-1}o{g}^{-1}\right)\left(x\right)=f^{-1}\left(g^{-1}\left(x\right)\right) \\ \Rightarrow \left(f^{-1}o{g}^{-1}\right)\left(x\right)=f^{-1}\left(x-2\right) \\ \Rightarrow \left(f^{-1}o{g}^{-1}\right)\left(x\right)=\frac{x-2}{2}...\left(3\right) \\ \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right) \\ =g\left(2x\right) \\ =2x+2 \\ \text{Let}{\left(gof\right)}^{-1}\left(x\right)\text{=y ....}\left(4\right) \\ x=\left(gof\right)\left(y\right) \\ \Rightarrow x=2y+2 \\ \Rightarrow 2y=x-2 \\ \Rightarrow y=\frac{x-2}{2} \\ \Rightarrow {\left(gof\right)}^{-1}\left(x\right)=\frac{x-2}{2}\text{[from}\left(4\right)...\left(5\right)] \\ \text{From}\left(3\right)\text{and}\left(5\right)\text{,} \\ {\left(gof\right)}^{-1}=f^{-1}o{g}^{-1} \end{gathered}$$