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To find the composition [tex]\((f \cdot g)(x)\)[/tex] where [tex]\( f(x) = \frac{x}{x-3} \)[/tex] and [tex]\( g(x) = \frac{1}{x} \)[/tex]:
1. Find the composition [tex]\( (f \cdot g)(x) \)[/tex]:
[tex]\[ (f \cdot g)(x) = f(g(x)) = f\left(\frac{1}{x}\right) \][/tex]
Substitute [tex]\( g(x) = \frac{1}{x} \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f\left(\frac{1}{x}\right) = \frac{\frac{1}{x}}{\frac{1}{x} - 3} \][/tex]
2. Simplify the composite function:
[tex]\[ f\left(\frac{1}{x}\right) = \frac{\frac{1}{x}}{\frac{1 - 3x}{x}} = \frac{\frac{1}{x}}{\frac{1 - 3x}{x}} = \frac{1}{1-3x} \][/tex]
Further simplifying the result, we get:
[tex]\[ f\left(\frac{1}{x}\right) = -\frac{1}{3x - 1} \][/tex]
3. Determine the domain of the composite function [tex]\( f \cdot g \)[/tex]:
- The function [tex]\( g(x) = \frac{1}{x} \)[/tex] is defined for all [tex]\( x \neq 0 \)[/tex]. Thus [tex]\( x \neq 0 \)[/tex].
- The function [tex]\( f(x) = \frac{x}{x-3} \)[/tex] is defined for all [tex]\( x \neq 3 \)[/tex]. However, since we are substituting [tex]\( g(x) = \frac{1}{x} \)[/tex] into it, the value [tex]\( x \)[/tex] that makes the denominator of [tex]\( f(g(x)) \)[/tex] zero must be avoided. Hence, for [tex]\( \frac{1}{x} - 3 \neq 0 \)[/tex]:
[tex]\[ \frac{1}{x} \neq 3 \Rightarrow x \neq \frac{1}{3} \][/tex]
Combining these two conditions, the domain of the composite function [tex]\( (f \cdot g)(x) \)[/tex] is:
[tex]\[ x \neq 0 \text{ and } x \neq \frac{1}{3} \][/tex]
So, in interval notation, the domain is:
[tex]\[ (-\infty, 0) \cup (0, \frac{1}{3}) \cup (\frac{1}{3}, \infty) \][/tex]
So, the composition and its domain are as follows:
[tex]\[ (f \cdot g)(x) = -\frac{1}{3x - 1} \][/tex]
[tex]\[ \text{Domain of } (f \cdot g): (-\infty, 0) \cup (0, 3) \cup (3, \infty) \][/tex]
1. Find the composition [tex]\( (f \cdot g)(x) \)[/tex]:
[tex]\[ (f \cdot g)(x) = f(g(x)) = f\left(\frac{1}{x}\right) \][/tex]
Substitute [tex]\( g(x) = \frac{1}{x} \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f\left(\frac{1}{x}\right) = \frac{\frac{1}{x}}{\frac{1}{x} - 3} \][/tex]
2. Simplify the composite function:
[tex]\[ f\left(\frac{1}{x}\right) = \frac{\frac{1}{x}}{\frac{1 - 3x}{x}} = \frac{\frac{1}{x}}{\frac{1 - 3x}{x}} = \frac{1}{1-3x} \][/tex]
Further simplifying the result, we get:
[tex]\[ f\left(\frac{1}{x}\right) = -\frac{1}{3x - 1} \][/tex]
3. Determine the domain of the composite function [tex]\( f \cdot g \)[/tex]:
- The function [tex]\( g(x) = \frac{1}{x} \)[/tex] is defined for all [tex]\( x \neq 0 \)[/tex]. Thus [tex]\( x \neq 0 \)[/tex].
- The function [tex]\( f(x) = \frac{x}{x-3} \)[/tex] is defined for all [tex]\( x \neq 3 \)[/tex]. However, since we are substituting [tex]\( g(x) = \frac{1}{x} \)[/tex] into it, the value [tex]\( x \)[/tex] that makes the denominator of [tex]\( f(g(x)) \)[/tex] zero must be avoided. Hence, for [tex]\( \frac{1}{x} - 3 \neq 0 \)[/tex]:
[tex]\[ \frac{1}{x} \neq 3 \Rightarrow x \neq \frac{1}{3} \][/tex]
Combining these two conditions, the domain of the composite function [tex]\( (f \cdot g)(x) \)[/tex] is:
[tex]\[ x \neq 0 \text{ and } x \neq \frac{1}{3} \][/tex]
So, in interval notation, the domain is:
[tex]\[ (-\infty, 0) \cup (0, \frac{1}{3}) \cup (\frac{1}{3}, \infty) \][/tex]
So, the composition and its domain are as follows:
[tex]\[ (f \cdot g)(x) = -\frac{1}{3x - 1} \][/tex]
[tex]\[ \text{Domain of } (f \cdot g): (-\infty, 0) \cup (0, 3) \cup (3, \infty) \][/tex]
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