If f : Q → Q, g : Q → Q are two functions defined by f(x) = 2 x and g(x) = x + 2,

Solution:

Injectivity of $f$.

Let $x$ and $y$ be two elements of domain $(Q)$, such that $f(x)=f(y)$

$\Rightarrow 2 x=2 y$

$\Rightarrow x=y$

So, $f$ is one-one.

 

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

$\Rightarrow 2 x=y$

$\Rightarrow x=\frac{y}{2} \in Q \quad$ (domain)

$\Rightarrow f$ is onto.

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

Finding $f^{-1}$ :

Let $f^{-1}(x)=y \quad \ldots(1)$

$\Rightarrow x=f(y)$

$\Rightarrow x=2 y$

$\Rightarrow y=\frac{x}{2}$

So, $f^{-1}(x)=\frac{x}{2}$                 (from (1))

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.

$\Rightarrow x+2=y$

$\Rightarrow x=2-y \in Q \quad$ (domain)

$\Rightarrow g$ is onto.

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

Finding $g^{-1}$ :

Let $g^{-1}(x)=y \quad \ldots(2)$

$\Rightarrow x=g(y)$

$\Rightarrow x=y+2$

$\Rightarrow y=x-2$

So, $g^{-1}(x)=x-2$                  (From (2))

 

Verification of $(g \circ f)^{-1}=f^{-1} o g^{-1}$ :

$f(x)=2 x ; g(x)=x+2$

and $f^{-1}(x)=\frac{x}{2} ; g^{-1}(x)=x-2$

Now, $\left(f^{-1} o g^{-1}\right)(x)=f^{-1}\left(g^{-1}(x)\right)$

$\Rightarrow\left(f^{-1} o g^{-1}\right)(x)=f^{-1}(x-2)$

$\Rightarrow\left(f^{-1} o g^{-1}\right)(x)=\frac{x-2}{2} \quad \ldots(3)$

$(g o f)(x)=g(f(x))$

$=g(2 x)$

$=2 x+2$

Let $(g o f)^{-1}(x)=\mathrm{y} \ldots$

$x=(g o f)(y)$

$\Rightarrow x=2 y+2$

$\Rightarrow 2 y=x-2$

$\Rightarrow y=\frac{x-2}{2}$

$\Rightarrow(g o f)^{-1}(x)=\frac{x-2}{2} \quad[$ from (4) $\ldots(5)]$

From (3) and (5),

$(g o f)^{-1}=f^{-1} o g^{-1}$