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To determine the domain of the composite function [tex]\((f \circ g)(x)\)[/tex], where [tex]\(f(x) = \frac{1}{x+5}\)[/tex] and [tex]\(g(x) = \frac{5}{-2x + 9}\)[/tex], we need to follow a series of detailed steps. Let’s go through them.
1. Determine the Domain of [tex]\(g(x)\)[/tex]:
The function [tex]\(g(x) = \frac{5}{-2x + 9}\)[/tex] is defined for all [tex]\(x\)[/tex] except where the denominator is zero. So we need to solve for [tex]\(x\)[/tex] in [tex]\(-2x + 9 = 0\)[/tex].
[tex]\[ -2x + 9 = 0 \][/tex]
[tex]\[ -2x = -9 \][/tex]
[tex]\[ x = \frac{9}{2} \][/tex]
Therefore, [tex]\(g(x)\)[/tex] is undefined at [tex]\(x = \frac{9}{2}\)[/tex]. The domain of [tex]\(g(x)\)[/tex] is:
[tex]\[ (-\infty, \frac{9}{2}) \cup (\frac{9}{2}, \infty) \][/tex]
2. Determine the Range of [tex]\(g(x)\)[/tex]:
The function [tex]\(g(x) = \frac{5}{-2x + 9}\)[/tex] can take on any real value except where the expression inside [tex]\(f\)[/tex] causes [tex]\(f(x)\)[/tex] to be undefined. We know that [tex]\(f(x) = \frac{1}{x+5}\)[/tex] is undefined at [tex]\(x = -5\)[/tex].
3. Determine when [tex]\(g(x)\)[/tex] equals [tex]\(-5\)[/tex]:
We need to find when [tex]\(-2x + 9 = -5\)[/tex]. So we solve:
[tex]\[ -2x + 9 = -5 \][/tex]
[tex]\[ -2x = -5 - 9 \][/tex]
[tex]\[ -2x = -14 \][/tex]
[tex]\[ x = 7 \][/tex]
Therefore, [tex]\(g(x) = -5\)[/tex] when [tex]\(x = 7\)[/tex]. Since [tex]\(f(g(x))\)[/tex] must avoid situations where [tex]\(g(x) = -5\)[/tex], this value needs to be excluded from the domain of the composite function.
4. Combine the Restrictions:
We must exclude both [tex]\(x = \frac{9}{2}\)[/tex] and [tex]\(x = 7\)[/tex] from the domain. The combined domain must account for these values:
[tex]\[ (-\infty, 7) \cup (7, \frac{9}{2}) \cup (\frac{9}{2}, \infty) \][/tex]
To summarize, the domain of [tex]\((f \circ g)(x)\)[/tex] is:
[tex]\[ (-\infty, 7) \cup (7, 4.5) \cup (4.5, \infty) \][/tex]
Thus, the solution to the domain of [tex]\((f \circ g)(x)\)[/tex] incorporating the constraints from both [tex]\(g(x)\)[/tex] and [tex]\(f(g(x))\)[/tex] is:
[tex]\[ [(-\infty, 4.5), (4.5, \infty)], [(-\infty, 7), (7, 4.5), (4.5, \infty)] \][/tex]
1. Determine the Domain of [tex]\(g(x)\)[/tex]:
The function [tex]\(g(x) = \frac{5}{-2x + 9}\)[/tex] is defined for all [tex]\(x\)[/tex] except where the denominator is zero. So we need to solve for [tex]\(x\)[/tex] in [tex]\(-2x + 9 = 0\)[/tex].
[tex]\[ -2x + 9 = 0 \][/tex]
[tex]\[ -2x = -9 \][/tex]
[tex]\[ x = \frac{9}{2} \][/tex]
Therefore, [tex]\(g(x)\)[/tex] is undefined at [tex]\(x = \frac{9}{2}\)[/tex]. The domain of [tex]\(g(x)\)[/tex] is:
[tex]\[ (-\infty, \frac{9}{2}) \cup (\frac{9}{2}, \infty) \][/tex]
2. Determine the Range of [tex]\(g(x)\)[/tex]:
The function [tex]\(g(x) = \frac{5}{-2x + 9}\)[/tex] can take on any real value except where the expression inside [tex]\(f\)[/tex] causes [tex]\(f(x)\)[/tex] to be undefined. We know that [tex]\(f(x) = \frac{1}{x+5}\)[/tex] is undefined at [tex]\(x = -5\)[/tex].
3. Determine when [tex]\(g(x)\)[/tex] equals [tex]\(-5\)[/tex]:
We need to find when [tex]\(-2x + 9 = -5\)[/tex]. So we solve:
[tex]\[ -2x + 9 = -5 \][/tex]
[tex]\[ -2x = -5 - 9 \][/tex]
[tex]\[ -2x = -14 \][/tex]
[tex]\[ x = 7 \][/tex]
Therefore, [tex]\(g(x) = -5\)[/tex] when [tex]\(x = 7\)[/tex]. Since [tex]\(f(g(x))\)[/tex] must avoid situations where [tex]\(g(x) = -5\)[/tex], this value needs to be excluded from the domain of the composite function.
4. Combine the Restrictions:
We must exclude both [tex]\(x = \frac{9}{2}\)[/tex] and [tex]\(x = 7\)[/tex] from the domain. The combined domain must account for these values:
[tex]\[ (-\infty, 7) \cup (7, \frac{9}{2}) \cup (\frac{9}{2}, \infty) \][/tex]
To summarize, the domain of [tex]\((f \circ g)(x)\)[/tex] is:
[tex]\[ (-\infty, 7) \cup (7, 4.5) \cup (4.5, \infty) \][/tex]
Thus, the solution to the domain of [tex]\((f \circ g)(x)\)[/tex] incorporating the constraints from both [tex]\(g(x)\)[/tex] and [tex]\(f(g(x))\)[/tex] is:
[tex]\[ [(-\infty, 4.5), (4.5, \infty)], [(-\infty, 7), (7, 4.5), (4.5, \infty)] \][/tex]
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