Math Problem Statement
Question content area top Part 1 Given f(x)equals7 plus StartFraction 1 Over x EndFraction and g(x)equalsStartFraction 1 Over x EndFraction find fplusg, fminusg, fg, and StartFraction f Over g EndFraction . Determine the domain for each function. Question content area bottom Part 1 (fplusg)(x)equals 7 plus StartFraction 2 Over x EndFraction (Simplify your answer.) Part 2 What is the domain of f plus g? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The domain of fplusg is StartSet nothing EndSet . (Use a comma to separate answers as needed.) B. The domain of fplusg is left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis. (Type your answer in interval notation.) Your answer is correct.C. The domain of fplusg is empty set. Part 3 (fminusg)(x)equals 7 (Simplify your answer.) Part 4 What is the domain of f minus g? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The domain of fminusg is StartSet nothing EndSet . (Use a comma to separate answers as needed.) B. The domain of fminusg is left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis. (Type your answer in interval notation.) Your answer is correct.C. The domain of fminusg is empty set. Part 5 (fg)(x)equals enter your response here (Simplify your answer.)
Solution
Let's break this problem down step by step. We're given two functions:
We need to find:
Part 1: Find
We can add the functions directly:
= \left( 7 + \frac{1}{x} \right) + \frac{1}{x} = 7 + \frac{2}{x}$$ So, $$(f + g)(x) = 7 + \frac{2}{x}$$. ### Part 2: Domain of $$(f + g)(x)$$ Both functions $$f(x)$$ and $$g(x)$$ involve division by $$x$$, so the domain will exclude $$x = 0$$. Hence, the domain of $$(f + g)(x)$$ is: $$(-\infty, 0) \cup (0, \infty)$$ ### Part 3: Find $$(f - g)(x)$$ Now, subtract the functions: $$(f - g)(x) = f(x) - g(x) = \left( 7 + \frac{1}{x} \right) - \frac{1}{x} = 7$$ So, $$(f - g)(x) = 7$$. ### Part 4: Domain of $$(f - g)(x)$$ Since $$(f - g)(x) = 7$$ is a constant and does not involve $$x$$, its domain is the same as the domain of the original functions, which is: $$(-\infty, 0) \cup (0, \infty)$$ ### Part 5: Find $$(fg)(x)$$ Now, multiply the functions: $$(fg)(x) = f(x) \cdot g(x) = \left( 7 + \frac{1}{x} \right) \cdot \frac{1}{x} = \frac{7}{x} + \frac{1}{x^2}$$ So, $$(fg)(x) = \frac{7}{x} + \frac{1}{x^2}$$. ### Part 6: Find $$\frac{f}{g}(x)$$ Finally, divide the functions: $$\frac{f}{g}(x) = \frac{f(x)}{g(x)} = \frac{7 + \frac{1}{x}}{\frac{1}{x}} = \left(7 + \frac{1}{x}\right) \cdot x = 7x + 1$$ So, $$\frac{f}{g}(x) = 7x + 1$$. ### Summary: 1. $$(f + g)(x) = 7 + \frac{2}{x}$$ 2. Domain of $$(f + g)(x)$$ is $$(-\infty, 0) \cup (0, \infty)$$ 3. $$(f - g)(x) = 7$$ 4. Domain of $$(f - g)(x)$$ is $$(-\infty, 0) \cup (0, \infty)$$ 5. $$(fg)(x) = \frac{7}{x} + \frac{1}{x^2}$$ 6. $$\frac{f}{g}(x) = 7x + 1$$ Would you like further explanation on any of these steps or the domains? --- Here are some related questions for practice: 1. How would you find the domain of the function $$f(x) = \frac{1}{x - 3}$$? 2. What happens to the domain of $$f(x) = \frac{1}{x - 3}$$ if we multiply by $$g(x) = x + 2$$? 3. Can you find the domain of $$\frac{1}{x^2 - 9}$$? 4. What is the inverse of $$f(x) = 7 + \frac{1}{x}$$? 5. How do you handle limits involving functions with denominators like these? **Tip:** Always pay close attention to points where division by zero might occur, as these will affect the domain of the function.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Functions
Domain of functions
Addition and subtraction of functions
Multiplication and division of functions
Formulas
f(x) + g(x) = f(x) + g(x)
f(x) - g(x) = f(x) - g(x)
f(x) * g(x) = f(x) * g(x)
f(x) / g(x) = f(x) / g(x)
Theorems
Domain of a rational function
Domain restrictions due to division by zero
Suitable Grade Level
Grades 9-11
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