Consider the following function:
\(\begin{array}{c}g\left( x \right)= 7x - 11\\h\left( x \right)= \frac{1}{7}x + 11\end{array}\)
Now, find \(\left( {h \circ g} \right)\left( x \right)\) as follow;
\(\begin{array}{c}\left( {h \circ g} \right)\left( x \right)= h\left( x \right)\\ = h\left( {7x - 11} \right)\\
= \frac{1}{7}\left( {7x - 11} \right) + 11\\= x - \frac{{11}}{7} + 11\end{array}\)
Therefore,
\(\left( {h \circ g} \right)\left( x \right) = x + \frac{{66}}{7}\)
Now, find \(\left( {g \circ h} \right)\left( x \right)\) as follow:
\(\begin{array}{c}\left( {g \circ h} \right)\left( x \right) = g\left( {h\left( x \right)} \right)\\ = g\left( {\frac{1}{7}x + 11} \right)\\ = 7\left( {\frac{1}{7}x + 11} \right) - 11\\ = x + 77 - 11\\\left( {g \circ h} \right)\left( x \right) = x + 66\end{array}\)
Since, from above it can see that,
\(\left( {h \circ g} \right)\left( x \right) \ne \left( {g \circ h} \right)\left( x \right)\)
Therefore, the above functions are not inverse functions.
\(\begin{array}{c}g\left( x \right)= 7x - 11\\h\left( x \right)= \frac{1}{7}x + 11\end{array}\)
Now, find \(\left( {h \circ g} \right)\left( x \right)\) as follow;
\(\begin{array}{c}\left( {h \circ g} \right)\left( x \right)= h\left( x \right)\\ = h\left( {7x - 11} \right)\\
= \frac{1}{7}\left( {7x - 11} \right) + 11\\= x - \frac{{11}}{7} + 11\end{array}\)
Therefore,
\(\left( {h \circ g} \right)\left( x \right) = x + \frac{{66}}{7}\)
Now, find \(\left( {g \circ h} \right)\left( x \right)\) as follow:
\(\begin{array}{c}\left( {g \circ h} \right)\left( x \right) = g\left( {h\left( x \right)} \right)\\ = g\left( {\frac{1}{7}x + 11} \right)\\ = 7\left( {\frac{1}{7}x + 11} \right) - 11\\ = x + 77 - 11\\\left( {g \circ h} \right)\left( x \right) = x + 66\end{array}\)
Since, from above it can see that,
\(\left( {h \circ g} \right)\left( x \right) \ne \left( {g \circ h} \right)\left( x \right)\)
Therefore, the above functions are not inverse functions.