For the real-valued functions $$ f(x)=\frac{x-6}{x-5} $$ and $$ g(x)=4 x+9 $$, find the composition $$ f \circ g $$ and specify its domain using interval notation.

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The composition $$ f \circ g $$ is $$ \frac{4x + 3}{4x + 4} $$ and its domain is $$ (-\infty, -1) \cup (-1, \infty) $$. To find the composition $$ f \circ g $$, we need to evaluate $$ f(g(x)) $$. First, we have the functions: $$ f(x) = \frac{x-6}{x-5} $$ $$ g(x) = 4x + 9 $$ Now, we will substitute $$ g(x) $$ into $$ f(x) $$: $$ f(g(x)) = f(4x + 9) = \frac{(4x + 9) - 6}{(4x + 9) - 5} $$ Simplifying the numerator and the denominator: $$ = \frac{4x + 9 - 6}{4x + 9 - 5} = \frac{4x + 3}{4x + 4} $$ Thus, we have: $$ f \circ g (x) = \frac{4x + 3}{4x + 4} $$ Next, we need to determine the domain of $$ f \circ g $$. The function $$ f(x) $$ is undefined when its denominator is zero. Therefore, we need to find when $$ g(x) - 5 = 0 $$: $$ g(x) - 5 = 4x + 9 - 5 = 4x + 4 = 0 $$ Solving for $$ x $$: $$ 4x + 4 = 0 \implies 4x = -4 \implies x = -1 $$ Thus, $$ f(g(x)) $$ is undefined at $$ x = -1 $$. The domain of $$ f \circ g $$ is all real numbers except $$ x = -1 $$. In interval notation, the domain is: $$ (-\infty, -1) \cup (-1, \infty) $$ In summary, the composition $$ f \circ g $$ is: $$ f \circ g (x) = \frac{4x + 3}{4x + 4} $$ and its domain is: $$ (-\infty, -1) \cup (-1, \infty) $$

For the real-valued functions \( f(x)=\frac{x-6}{x-5} \) and \( g(x)=4 x+9 \), find the composition \( f \circ g \) and specify its domain using interval notation.

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For the real-valued functions \( f(x)=\frac{x-6}{x-5} \) and \( g(x)=4 x+9 \), find the composition \( f \circ g \) and specify its domain using interval notation.

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