Let f: R - {-frac{1}{2}} rightarrow R and g: | vedclass

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Question

(A) The domain of $f \circ g$ consists of all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$.$1$. The domain of $g$ is $R - \{-\fra

Vedclass · Accepted Answer

$R - \{-\frac{5}{2}\}$

Vedclass · Answer

(A) The domain of $f \circ g$ consists of all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$.$1$. The domain of $g$ is $R - \{-\frac{5}{2}\}$.$2$. For $f(g(x))$ to be defined,$g(x)$ must not be equal to $-\frac{1}{2}$ (the value excluded from the domain of $f$).$3$. Set $g(x) = -\frac{1}{2}$:$\frac{|x|+1}{2x+5} = -\frac{1}{2}$$2(|x|+1) = -(2x+5)$$2|x| + 2 = -2x - 5$$2|x| = -2x - 7$Case $I$: If $x \ge 0$,then $2x = -2x - 7 \Rightarrow 4x = -7 \Rightarrow x = -\frac{7}{4}$. Since $x \ge 0$,this is not a solution.Case $II$: If $x Thus,$g(x)$ is never equal to $-\frac{1}{2}$.Therefore,the only restriction on the domain of $f \circ g$ is $x 
eq -\frac{5}{2}$.The domain is $R - \{-\frac{5}{2}\}$.

Let $f: R - \{-\frac{1}{2}\} \rightarrow R$ and $g: R - \{-\frac{5}{2}\} \rightarrow R$ be defined as $f(x) = \frac{2x+3}{2x+1}$ and $g(x) = \frac{|x|+1}{2x+5}$. Then the domain of the function $f \circ g$ is:

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