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Sagot :
To find the domain of the composite function [tex]\((f \circ g)(x)\)[/tex], where [tex]\(f(x) = \frac{1}{x-1}\)[/tex] and [tex]\(g(x) = \frac{1}{x+4}\)[/tex], we need to follow these steps:
1. Find the domain of [tex]\(g(x)\)[/tex]:
- [tex]\(g(x)\)[/tex] is defined as [tex]\(\frac{1}{x+4}\)[/tex].
- The function [tex]\(g(x)\)[/tex] will be undefined when the denominator is zero, i.e., [tex]\(x + 4 = 0\)[/tex] which gives [tex]\(x = -4\)[/tex].
- So, the domain of [tex]\(g(x)\)[/tex] is all real numbers except [tex]\(x = -4\)[/tex], which can be expressed as:
[tex]\[ (-\infty, -4) \cup (-4, \infty) \][/tex]
2. Find the composite function [tex]\((f \circ g)(x)\)[/tex]:
- The composite function [tex]\((f \circ g)(x)\)[/tex] is given by [tex]\(f(g(x))\)[/tex].
- Substituting [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex], we get:
[tex]\[ (f \circ g)(x) = f\left(\frac{1}{x+4}\right) \][/tex]
- Next, we substitute [tex]\(\frac{1}{x+4}\)[/tex] into [tex]\(f(x) = \frac{1}{x-1}\)[/tex]:
[tex]\[ f\left(\frac{1}{x+4}\right) = \frac{1}{\left(\frac{1}{x+4}\right) - 1} = \frac{1}{\frac{1 - (x+4)}{x+4}} = \frac{1}{\frac{-x-3}{x+4}} = \frac{x+4}{-x-3} \][/tex]
3. Determine where [tex]\( (f \circ g)(x) \)[/tex] is undefined:
- [tex]\((f \circ g)(x)\)[/tex] will be undefined where [tex]\(g(x)\)[/tex] is undefined, i.e., at [tex]\(x = -4\)[/tex].
- In addition, [tex]\((f \circ g)(x)\)[/tex] will be undefined where [tex]\(f(g(x))\)[/tex] is undefined:
- [tex]\(f(x) = \frac{1}{x-1}\)[/tex] is undefined when [tex]\(x = 1\)[/tex].
- Solve [tex]\(g(x) = 1\)[/tex]:
[tex]\[ \frac{1}{x+4} = 1 \implies x+4 = 1 \implies x = -3 \][/tex]
- Hence, [tex]\((f \circ g)(x)\)[/tex] is undefined at [tex]\(x = -3\)[/tex].
4. Combine the results:
- The domain of [tex]\((f \circ g)(x)\)[/tex] combines the restrictions from [tex]\(g(x)\)[/tex] and the places where the composite function is undefined.
- Therefore, the domain of [tex]\((f \circ g)(x)\)[/tex] is:
[tex]\[ (-\infty, -4) \cup (-4, -3) \cup (-3, \infty) \][/tex]
Thus, the correct answer is:
[tex]\[ (-\infty, -4) \cup (-4, -3) \cup (-3, \infty) \][/tex]
1. Find the domain of [tex]\(g(x)\)[/tex]:
- [tex]\(g(x)\)[/tex] is defined as [tex]\(\frac{1}{x+4}\)[/tex].
- The function [tex]\(g(x)\)[/tex] will be undefined when the denominator is zero, i.e., [tex]\(x + 4 = 0\)[/tex] which gives [tex]\(x = -4\)[/tex].
- So, the domain of [tex]\(g(x)\)[/tex] is all real numbers except [tex]\(x = -4\)[/tex], which can be expressed as:
[tex]\[ (-\infty, -4) \cup (-4, \infty) \][/tex]
2. Find the composite function [tex]\((f \circ g)(x)\)[/tex]:
- The composite function [tex]\((f \circ g)(x)\)[/tex] is given by [tex]\(f(g(x))\)[/tex].
- Substituting [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex], we get:
[tex]\[ (f \circ g)(x) = f\left(\frac{1}{x+4}\right) \][/tex]
- Next, we substitute [tex]\(\frac{1}{x+4}\)[/tex] into [tex]\(f(x) = \frac{1}{x-1}\)[/tex]:
[tex]\[ f\left(\frac{1}{x+4}\right) = \frac{1}{\left(\frac{1}{x+4}\right) - 1} = \frac{1}{\frac{1 - (x+4)}{x+4}} = \frac{1}{\frac{-x-3}{x+4}} = \frac{x+4}{-x-3} \][/tex]
3. Determine where [tex]\( (f \circ g)(x) \)[/tex] is undefined:
- [tex]\((f \circ g)(x)\)[/tex] will be undefined where [tex]\(g(x)\)[/tex] is undefined, i.e., at [tex]\(x = -4\)[/tex].
- In addition, [tex]\((f \circ g)(x)\)[/tex] will be undefined where [tex]\(f(g(x))\)[/tex] is undefined:
- [tex]\(f(x) = \frac{1}{x-1}\)[/tex] is undefined when [tex]\(x = 1\)[/tex].
- Solve [tex]\(g(x) = 1\)[/tex]:
[tex]\[ \frac{1}{x+4} = 1 \implies x+4 = 1 \implies x = -3 \][/tex]
- Hence, [tex]\((f \circ g)(x)\)[/tex] is undefined at [tex]\(x = -3\)[/tex].
4. Combine the results:
- The domain of [tex]\((f \circ g)(x)\)[/tex] combines the restrictions from [tex]\(g(x)\)[/tex] and the places where the composite function is undefined.
- Therefore, the domain of [tex]\((f \circ g)(x)\)[/tex] is:
[tex]\[ (-\infty, -4) \cup (-4, -3) \cup (-3, \infty) \][/tex]
Thus, the correct answer is:
[tex]\[ (-\infty, -4) \cup (-4, -3) \cup (-3, \infty) \][/tex]
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