# If f(x) = x/6 - 2 and g(x) = 6x + 12, how can I show (f@g)(x) = (g@f)(x)?

##### 1 Answer
Jun 26, 2018

$\text{see explanation}$

#### Explanation:

$\text{if "(f@g)(x)=(g@f)(x)=x" then}$

$f \left(x\right) \text{ and "g(x)" are inverse functions of each other}$

$\text{if we show that "f(x)" and "g(x)" are inverse of each }$
$\text{other then the above statement should be true}$

$\text{let "y=x/6-2" and } y = 6 x + 12$

$\text{rearrange both making x the subject}$

$6 y = x - 12 \Rightarrow x = 6 y + 12$

${f}^{-} 1 \left(x\right) = 6 x + 12 = g \left(x\right)$

$y = 6 x + 12 \Rightarrow x = \frac{1}{6} \left(y - 12\right) = \frac{y}{6} - 2$

${g}^{-} 1 x = \frac{x}{6} - 2 = f \left(x\right)$

$\Rightarrow \left(f \circ g\right) \left(x\right) = \left(g \circ f\right) \left(x\right)$

$\textcolor{b l u e}{\text{As a check}}$

$\left(f \circ g\right) \left(x\right) = f \left(6 x + 12\right)$

$\textcolor{w h i t e}{\times \times \times x} = \frac{6 x + 12}{6} - 2 = x + 2 - 2 = x$

$\left(g \circ f\right) \left(x\right) = g \left(\frac{x}{6} - 2\right)$

$\textcolor{w h i t e}{\times \times \times x} = 6 \left(\frac{x}{6} - 2\right) + 12 = x - 12 + 12 = x$

$\Rightarrow \left(f \circ g\right) \left(x\right) = \left(g \circ f\right) \left(x\right)$