Question

Find f(g(x)). State the domain of the composite function. f(x) = 3x-2/x+1 g(x) = x+5/2x-3

Answer 1

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,

$g(x)=\dfrac{x+5}{2x-3}\Longrightarrow 2x-3\neq 0\iff \boxed{x\neq\dfrac{3}{2}}$

- Domain of f(g(x)): We'll find its expression:

$f(x) = \dfrac{3x-2}{x+1}\\\\f(g(x)) = \dfrac{3g(x)-2}{g(x)+1}\\\\f(g(x)) = \dfrac{3\cdot\dfrac{x+5}{2x-3}-2}{\dfrac{x+5}{2x-3}+1}=\dfrac{~~~\dfrac{3(x+5)-2(2x-3)}{2x-3}~~~}{\dfrac{(x+5)+(2x-3)}{2x-3}}\\\\f(g(x)) =\dfrac{3(x+5)-2(2x-3)}{(x+5)+(2x-3)}=\dfrac{3x+15-4x+6}{x+5+2x-3}\\\\\boxed{f(g(x)) =\dfrac{21-x}{3x+2}}$

Now, once again, we have to exclude the values of x that make the denominator equals to zero. Thus,

$f(g(x)) =\dfrac{21-x}{3x+2}\Longrightarrow 3x+2\neq0\iff \boxed{x\neq-\dfrac{2}{3}}$

Lastly, we may write the domanin of f(g(x)):

$D(f(g(x)) = \left]-\infty,-\dfrac{2}{3}\right[\cup\left]-\dfrac{2}{3},\dfrac{3}{2}\right[\cup\left]\dfrac{3}{2},\infty\right[$

or, just writing in a shorter way:

$\boxed{D(f(g(x)) = \mathbb{R}-\left\{-\dfrac{2}{3},\dfrac{3}{2}\right\}}$