Answer:
[tex]h(x)=\frac{x+1}{5x+7},Domain=All\text{ }Real\text{ }numbers,\text{ }except\text{ }x=-\frac{3}{2}\text{ }and\text{ }x=-\frac{7}{5}[/tex][tex]h^{-1}(x)=\frac{1-7x}{5x-1},Domain=All\text{ }Real\text{ }numbers,\text{ }except\text{ }x=\frac{1}{5}[/tex]
Explanation:
The notation for composition of functions is:
[tex](f\circ g)(x)=f(g(x))[/tex]In this case:
[tex]\begin{cases}f(x)={\frac{x}{x+2}} \\ g(x)={\frac{x+1}{2x+3}}\end{cases}[/tex]To do the composition, we replace the x in the f(x) with the function g(x):
[tex](f\circ g)(x)=f(g(x))=\frac{g(x)}{g(x)+3}=\frac{\frac{x+1}{2x+3}}{\frac{x+1}{2x+3}+2}[/tex]And solve:
[tex]=\frac{\frac{x+1}{2x+3}}{\frac{x+1}{2x+3}+2}=\frac{\frac{x+1}{2x+3}}{\frac{x+1}{2x+3}+\frac{2(2x+3)}{2x+3}}=\frac{\frac{x+1}{2x+3}}{\frac{5x+7}{2x+3}}=\frac{(x+1)(2x+3)}{(2x+3)(5x+7)}[/tex]Here, we can calcualte the domain. The function is not defined when teh denominator is 0, thus:
[tex]2x+3=0\Rightarrow x=-\frac{3}{2}[/tex][tex]5x+7=0\Rightarrow x=-\frac{7}{5}[/tex]Since the function can't be evaluated when x = -3/2, we can cancel the terms (2x+3) in the numerator and denominator:
[tex]\frac{(x+1)(2x+3)}{(2x+3)(5x+7)}=\frac{x+1}{5x+7}[/tex]Thus:
[tex]\begin{equation*} h(x)=\frac{x+1}{5x+7},Domain=All\text{ }Real\text{ }numbers,\text{ }except\text{ }x=-\frac{3}{2}\text{ }and\text{ }x=-\frac{7}{5} \end{equation*}[/tex]Now, to find the inverse of the function, we first switch the variables:
[tex]y=\frac{x+1}{5x+7}\Rightarrow x=\frac{y+1}{5y+7}[/tex]And solve for y:
[tex]\begin{gathered} \begin{equation*} x=\frac{y+1}{5y+7} \end{equation*} \\ . \\ x(5y+7)=y+1 \\ . \\ 5xy+7x=y+1 \\ . \\ 5xy-y=1-7x \\ . \\ y(5x-1)=1-7x \\ . \\ y=\frac{1-7x}{5x-1}\Rightarrow h^{-1}(x)=\frac{1-7x}{5x-1} \end{gathered}[/tex]And since the denominator can't be 0:
[tex]5x-1=0\Rightarrow x=\frac{1}{5}[/tex]Thus:
[tex]\begin{equation*} h^{-1}(x)=\frac{1-7x}{5x-1},Domain=All\text{ }Real\text{ }numbers,\text{ }except\text{ }x=\frac{1}{5} \end{equation*}[/tex]