$(f\circ g)(x) = f(g(x))\\\\(f\circ g)(x) = f\left(\dfrac{13}{x+3}\right)\\\\(f\circ g)(x) = \dfrac{2}{\left(\dfrac{13}{x+3}\right)+3}\\\\(f\circ g)(x) = \dfrac{2}{~~\dfrac{13+3(x+3)}{x+3}~~}\\\\(f\circ g)(x) = \dfrac{2(x+3)}{13+3(x+3)}\\\\\boxed{(f\circ g)(x) = \dfrac{2x+6}{3x+22}}$

Since we have a fraction with x in the denominator, we have to remove from the domain the value of that makes null the denominator. Thus,

$3x+22\neq0\Longrightarrow 3x\neq22\Longrightarrow x\neq \dfrac{22}{3}$

Furthermore, we also have to remove the value of x that annuls the denominator of g(x), because that's the first function that we use. So:

$x+3\neq0\Longrightarrow x\neq-3$

Hence, the domain is:

$D((f\circ g)(x)) = \left]-\infty,-3\right[\cup \left]-3,\dfrac{22}{3}\right[\cup\left]\dfrac{22}{3},\infty\right[$

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3 years ago

Divide. ​ 5 1/4÷1/3 ​ Express your answer as a mixed number in simplest form. Enter your answer in the box.

Divide. 5 1/4÷1/3 Express your answer as a mixed number in simplest form. Enter your answer in the box.