For the real-valued functions g(x)=(x-2)/(x+5) and h(x)=2x-11, find the composition g@h and specify its domain using interval notation.

Calculate composition g(h(x)): Step 1: Calculate the composition g(h(x)). g(x) = (x-2)/(x+5), h(x) = 2x-11. Substitute h(x) into g(x): g(h(x)) = g(2x-11) = ((2x-11)-2)/((2x-11)+5). Simplify the expression: g(h(x)) = (2x-13)/(2x-6).Determine domain of g(h(x)): Step 2: Determine the domain of g(h(x)). The domain of g(h(x)) = (2x-13)/(2x-6) is all real numbers except where the denominator is zero. Set the denominator equal to zero and solve for x: 2x - 6 = 0, 2x = 6, x = 3. The function is undefined at x = 3.

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For the real-valued functions $g(x)=\frac{x-2}{x+5}$ and $h(x)=2x-11$ , find the composition $g \circ h$ and specify its domain using interval notation.

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Algebra 1

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Q. For the real-valued functions

g(x)=\frac{x-2}{x+5}

and

h(x)=2x-11

, find the composition

g \circ h

and specify its domain using interval notation.

Calculate composition $g(h(x))$ : Step $1$ : Calculate the composition $g(h(x))$ . $g(x) = \frac{x-2}{x+5}$ , $h(x) = 2x-11$ . Substitute $h(x)$ into $g(x)$ : $g(h(x)) = g(2x-11) = \frac{(2x-11)-2}{(2x-11)+5}$ . Simplify the expression: $g(h(x)) = \frac{2x-13}{2x-6}$ .
Determine domain of $g(h(x))$ : Step $2$ : Determine the domain of $g(h(x))$ . The domain of $g(h(x)) = \frac{2x-13}{2x-6}$ is all real numbers except where the denominator is zero. Set the denominator equal to zero and solve for $x$ : $2x - 6 = 0$ , $2x = 6$ , $x = 3$ . The function is undefined at $x = 3$ .