Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
To find the domain of the composite function [tex]\((f \circ g)(x)\)[/tex], we need to determine where both functions are defined and where the composite function itself is defined. Let's break this problem into steps.
1. Identify the domain of [tex]\(g(x)\)[/tex]:
The function [tex]\(g(x) = \frac{1}{-6x + 5}\)[/tex] is defined for all [tex]\(x\)[/tex] except where the denominator is zero. Set the denominator equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[ -6x + 5 = 0 \implies x = \frac{5}{6} \][/tex]
Therefore, [tex]\(g(x)\)[/tex] is defined for all [tex]\(x \neq \frac{5}{6}\)[/tex].
2. Identify the domain of [tex]\(f(g(x))\)[/tex]:
The function [tex]\(f(x) = \frac{1}{-10x - 7}\)[/tex] requires that [tex]\(x \neq -\frac{7}{10}\)[/tex] to be defined. Therefore, [tex]\(f\)[/tex] is defined for all [tex]\(x \neq -\frac{7}{10}\)[/tex]. In our case, we need this to [tex]\(f(g(x))\)[/tex].
To find where [tex]\(f(g(x))\)[/tex] is defined, we substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[ f(g(x)) = f\left( \frac{1}{-6x + 5} \right) = \frac{1}{-10 \left( \frac{1}{-6x + 5} \right) - 7} \][/tex]
3. Determine when [tex]\(f(g(x))\)[/tex] is undefined:
The function [tex]\(f(g(x))\)[/tex] becomes undefined when its denominator is zero:
[tex]\[ -10 \left( \frac{1}{-6x + 5} \right) - 7 = 0 \][/tex]
Simplify inside the denominator:
[tex]\[ - \frac{10}{-6x + 5} - 7 = 0 \][/tex]
Rearrange to solve for [tex]\(x\)[/tex]:
[tex]\[ - \frac{10}{-6x + 5} = 7 \implies \frac{10}{6x - 5} = 7 \implies 10 = 7(6x - 5) \implies 10 = 42x - 35 \implies 42x = 45 \implies x = \frac{45}{42} = \frac{15}{14} \][/tex]
Therefore, the domain of the composite function [tex]\((f \circ g)(x)\)[/tex] is all real numbers except where the original [tex]\(g(x)\)[/tex] or the intermediate steps of [tex]\(f(g(x))\)[/tex] become undefined.
Given the calculations:
- [tex]\(g(x)\)[/tex] is undefined at [tex]\(x = \frac{5}{6}}. - \(f(g(x))\)[/tex] is undefined at [tex]\(x = \frac{15}{14}}. Thus, the domain of \((f \circ g)(x)\)[/tex] is all real numbers except [tex]\(x = \frac{5}{6}\)[/tex] and [tex]\(x = \frac{15}{14}\)[/tex].
The correct answer is:
All real numbers except [tex]\(\frac{5}{6}\)[/tex] and [tex]\(\frac{15}{14} \)[/tex].
Therefore, the correct option is:
#### All real numbers except [tex]\(\frac{5}{6}\)[/tex] and [tex]\(\frac{15}{14}\)[/tex].
1. Identify the domain of [tex]\(g(x)\)[/tex]:
The function [tex]\(g(x) = \frac{1}{-6x + 5}\)[/tex] is defined for all [tex]\(x\)[/tex] except where the denominator is zero. Set the denominator equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[ -6x + 5 = 0 \implies x = \frac{5}{6} \][/tex]
Therefore, [tex]\(g(x)\)[/tex] is defined for all [tex]\(x \neq \frac{5}{6}\)[/tex].
2. Identify the domain of [tex]\(f(g(x))\)[/tex]:
The function [tex]\(f(x) = \frac{1}{-10x - 7}\)[/tex] requires that [tex]\(x \neq -\frac{7}{10}\)[/tex] to be defined. Therefore, [tex]\(f\)[/tex] is defined for all [tex]\(x \neq -\frac{7}{10}\)[/tex]. In our case, we need this to [tex]\(f(g(x))\)[/tex].
To find where [tex]\(f(g(x))\)[/tex] is defined, we substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[ f(g(x)) = f\left( \frac{1}{-6x + 5} \right) = \frac{1}{-10 \left( \frac{1}{-6x + 5} \right) - 7} \][/tex]
3. Determine when [tex]\(f(g(x))\)[/tex] is undefined:
The function [tex]\(f(g(x))\)[/tex] becomes undefined when its denominator is zero:
[tex]\[ -10 \left( \frac{1}{-6x + 5} \right) - 7 = 0 \][/tex]
Simplify inside the denominator:
[tex]\[ - \frac{10}{-6x + 5} - 7 = 0 \][/tex]
Rearrange to solve for [tex]\(x\)[/tex]:
[tex]\[ - \frac{10}{-6x + 5} = 7 \implies \frac{10}{6x - 5} = 7 \implies 10 = 7(6x - 5) \implies 10 = 42x - 35 \implies 42x = 45 \implies x = \frac{45}{42} = \frac{15}{14} \][/tex]
Therefore, the domain of the composite function [tex]\((f \circ g)(x)\)[/tex] is all real numbers except where the original [tex]\(g(x)\)[/tex] or the intermediate steps of [tex]\(f(g(x))\)[/tex] become undefined.
Given the calculations:
- [tex]\(g(x)\)[/tex] is undefined at [tex]\(x = \frac{5}{6}}. - \(f(g(x))\)[/tex] is undefined at [tex]\(x = \frac{15}{14}}. Thus, the domain of \((f \circ g)(x)\)[/tex] is all real numbers except [tex]\(x = \frac{5}{6}\)[/tex] and [tex]\(x = \frac{15}{14}\)[/tex].
The correct answer is:
All real numbers except [tex]\(\frac{5}{6}\)[/tex] and [tex]\(\frac{15}{14} \)[/tex].
Therefore, the correct option is:
#### All real numbers except [tex]\(\frac{5}{6}\)[/tex] and [tex]\(\frac{15}{14}\)[/tex].
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.