Answer:
ℝ - {(-2/3),(3/2)}
Step-by-step explanation:
We want the domain of f(g(x)). So, firstly, we have to find the domain for g(x) and, then, for f(g(x)).
- Domain of g(x): Since the expression is a fracion, we must exclude the values of x that make null the denominator. Hence,
[tex]g(x)=\dfrac{x+5}{2x-3}\Longrightarrow 2x-3\neq 0\iff \boxed{x\neq\dfrac{3}{2}}[/tex]
- Domain of f(g(x)): We'll find its expression:
[tex]f(x) = \dfrac{3x-2}{x+1}\\\\f(g(x)) = \dfrac{3g(x)-2}{g(x)+1}\\\\f(g(x)) = \dfrac{3\cdot\dfrac{x+5}{2x-3}-2}{\dfrac{x+5}{2x-3}+1}=\dfrac{~~~\dfrac{3(x+5)-2(2x-3)}{2x-3}~~~}{\dfrac{(x+5)+(2x-3)}{2x-3}}\\\\f(g(x)) =\dfrac{3(x+5)-2(2x-3)}{(x+5)+(2x-3)}=\dfrac{3x+15-4x+6}{x+5+2x-3}\\\\\boxed{f(g(x)) =\dfrac{21-x}{3x+2}}[/tex]
Now, once again, we have to exclude the values of x that make the denominator equals to zero. Thus,
[tex]f(g(x)) =\dfrac{21-x}{3x+2}\Longrightarrow 3x+2\neq0\iff \boxed{x\neq-\dfrac{2}{3}}[/tex]
Lastly, we may write the domanin of f(g(x)):
[tex]D(f(g(x)) = \left]-\infty,-\dfrac{2}{3}\right[\cup\left]-\dfrac{2}{3},\dfrac{3}{2}\right[\cup\left]\dfrac{3}{2},\infty\right[[/tex]
or, just writing in a shorter way:
[tex]\boxed{D(f(g(x)) = \mathbb{R}-\left\{-\dfrac{2}{3},\dfrac{3}{2}\right\}}[/tex]
