Question
For f left parenthesis x right parenthesis equals StartFraction 7 Over x plus 8 EndFraction
and g left parenthesis x right parenthesis equals StartFraction 7 Over x EndFraction
, find
a. left parenthesis f circle g right parenthesis left parenthesis x right parenthesis; b. the domain of f circle g
Question content area bottom
Part 1
a. left parenthesis f circle g right parenthesis left parenthesis x right parenthesisequals
enter your response here
(Simplify your answer.)
Part 2
b. What is the domain of f circle g?
The domain is
enter your response here.
(Simplify your answer. Type your answer in interval notation. Use integers or fractions for any numbers in the expression.)
and g left parenthesis x right parenthesis equals StartFraction 7 Over x EndFraction
, find
a. left parenthesis f circle g right parenthesis left parenthesis x right parenthesis; b. the domain of f circle g
Question content area bottom
Part 1
a. left parenthesis f circle g right parenthesis left parenthesis x right parenthesisequals
enter your response here
(Simplify your answer.)
Part 2
b. What is the domain of f circle g?
The domain is
enter your response here.
(Simplify your answer. Type your answer in interval notation. Use integers or fractions for any numbers in the expression.)
Answers
GPT-4o mini
Let's find \( (f \circ g)(x) \) and the domain of \( f \circ g \).
### Part 1: Finding \( (f \circ g)(x) \)
We have:
- \( f(x) = \frac{7}{x + 8} \)
- \( g(x) = \frac{7}{x} \)
To find \( (f \circ g)(x) \), we need to substitute \( g(x) \) into \( f(x) \):
\[
(f \circ g)(x) = f(g(x)) = f\left(\frac{7}{x}\right)
\]
Now, replace \( x \) in \( f(x) \) with \( g(x) \):
\[
f\left(\frac{7}{x}\right) = \frac{7}{\frac{7}{x} + 8}
\]
Now, simplify the denominator:
\[
\frac{7}{x} + 8 = \frac{7 + 8x}{x}
\]
So, we have:
\[
f\left(\frac{7}{x}\right) = \frac{7}{\frac{7 + 8x}{x}} = \frac{7x}{7 + 8x}
\]
Thus,
\[
(f \circ g)(x) = \frac{7x}{7 + 8x}
\]
### Part 2: Finding the Domain of \( f \circ g \)
To determine the domain of \( (f \circ g)(x) = \frac{7x}{7 + 8x} \), we need to identify values of \( x \) that make the denominator zero:
Set the denominator equal to zero and solve:
\[
7 + 8x = 0
\]
\[
8x = -7 \quad \Rightarrow \quad x = -\frac{7}{8}
\]
The function \( g(x) = \frac{7}{x} \) is also undefined when \( x = 0 \), as the denominator cannot be zero.
Thus, the domain of \( (f \circ g)(x) \) is all real numbers except \( x = -\frac{7}{8} \) and \( x = 0 \).
### In Interval Notation
The domain in interval notation can be expressed as:
\[
(-\infty, -\frac{7}{8}) \cup (-\frac{7}{8}, 0) \cup (0, \infty)
\]
### Final Answers
a. \( (f \circ g)(x) = \frac{7x}{7 + 8x} \)
b. The domain is \( (-\infty, -\frac{7}{8}) \cup (-\frac{7}{8}, 0) \cup (0, \infty) \)
### Part 1: Finding \( (f \circ g)(x) \)
We have:
- \( f(x) = \frac{7}{x + 8} \)
- \( g(x) = \frac{7}{x} \)
To find \( (f \circ g)(x) \), we need to substitute \( g(x) \) into \( f(x) \):
\[
(f \circ g)(x) = f(g(x)) = f\left(\frac{7}{x}\right)
\]
Now, replace \( x \) in \( f(x) \) with \( g(x) \):
\[
f\left(\frac{7}{x}\right) = \frac{7}{\frac{7}{x} + 8}
\]
Now, simplify the denominator:
\[
\frac{7}{x} + 8 = \frac{7 + 8x}{x}
\]
So, we have:
\[
f\left(\frac{7}{x}\right) = \frac{7}{\frac{7 + 8x}{x}} = \frac{7x}{7 + 8x}
\]
Thus,
\[
(f \circ g)(x) = \frac{7x}{7 + 8x}
\]
### Part 2: Finding the Domain of \( f \circ g \)
To determine the domain of \( (f \circ g)(x) = \frac{7x}{7 + 8x} \), we need to identify values of \( x \) that make the denominator zero:
Set the denominator equal to zero and solve:
\[
7 + 8x = 0
\]
\[
8x = -7 \quad \Rightarrow \quad x = -\frac{7}{8}
\]
The function \( g(x) = \frac{7}{x} \) is also undefined when \( x = 0 \), as the denominator cannot be zero.
Thus, the domain of \( (f \circ g)(x) \) is all real numbers except \( x = -\frac{7}{8} \) and \( x = 0 \).
### In Interval Notation
The domain in interval notation can be expressed as:
\[
(-\infty, -\frac{7}{8}) \cup (-\frac{7}{8}, 0) \cup (0, \infty)
\]
### Final Answers
a. \( (f \circ g)(x) = \frac{7x}{7 + 8x} \)
b. The domain is \( (-\infty, -\frac{7}{8}) \cup (-\frac{7}{8}, 0) \cup (0, \infty) \)