Question
Find left parenthesis f circle g right parenthesis left parenthesis x right parenthesis and its domain when f left parenthesis x right parenthesis equals x squared plus 2 and g left parenthesis x right parenthesis equals start root x plus 5 end root.
(1 point)
Responses
left parenthesis f circle g right parenthesis left parenthesis x right parenthesis equals start root x squared plus 7 end root semicolon left bracket negative 5 comma infinity right parenthesis
Image with alt text: left parenthesis f circle g right parenthesis left parenthesis x right parenthesis equals start root x squared plus 7 end root semicolon left bracket negative 5 comma infinity right parenthesis
left parenthesis f circle g right parenthesis left parenthesis x right parenthesis equals x plus 7 semicolon left bracket negative 5 comma infinity right parenthesis
Image with alt text: left parenthesis f circle g right parenthesis left parenthesis x right parenthesis equals x plus 7 semicolon left bracket negative 5 comma infinity right parenthesis
left parenthesis f circle g right parenthesis left parenthesis x right parenthesis equals x plus 7 semicolon left parenthesis negative infinity comma infinity right parenthesis
Image with alt text: left parenthesis f circle g right parenthesis left parenthesis x right parenthesis equals x plus 7 semicolon left parenthesis negative infinity comma infinity right parenthesis
left parenthesis f circle g right parenthesis left parenthesis x right parenthesis equals start root x squared plus 7 end root semicolon left parenthesis negative infinity comma infinity right parenthesis
(1 point)
Responses
left parenthesis f circle g right parenthesis left parenthesis x right parenthesis equals start root x squared plus 7 end root semicolon left bracket negative 5 comma infinity right parenthesis
Image with alt text: left parenthesis f circle g right parenthesis left parenthesis x right parenthesis equals start root x squared plus 7 end root semicolon left bracket negative 5 comma infinity right parenthesis
left parenthesis f circle g right parenthesis left parenthesis x right parenthesis equals x plus 7 semicolon left bracket negative 5 comma infinity right parenthesis
Image with alt text: left parenthesis f circle g right parenthesis left parenthesis x right parenthesis equals x plus 7 semicolon left bracket negative 5 comma infinity right parenthesis
left parenthesis f circle g right parenthesis left parenthesis x right parenthesis equals x plus 7 semicolon left parenthesis negative infinity comma infinity right parenthesis
Image with alt text: left parenthesis f circle g right parenthesis left parenthesis x right parenthesis equals x plus 7 semicolon left parenthesis negative infinity comma infinity right parenthesis
left parenthesis f circle g right parenthesis left parenthesis x right parenthesis equals start root x squared plus 7 end root semicolon left parenthesis negative infinity comma infinity right parenthesis
Answers
GPT-4o mini
To find \( (f \circ g)(x) \), we need to evaluate \( f(g(x)) \).
We have the following functions:
- \( f(x) = x^2 + 2 \)
- \( g(x) = \sqrt{x + 5} \)
Now, we compute \( (f \circ g)(x) \):
\[
g(x) = \sqrt{x + 5}
\]
Substituting \( g(x) \) into \( f(x) \):
\[
f(g(x)) = f(\sqrt{x + 5}) = (\sqrt{x + 5})^2 + 2
\]
Calculating this gives:
\[
(\sqrt{x + 5})^2 = x + 5
\]
Thus, we have:
\[
f(g(x)) = x + 5 + 2 = x + 7
\]
Now we need to determine the domain of \( (f \circ g)(x) \). The domain of \( g(x) \) must be considered first. The function \( g(x) = \sqrt{x + 5} \) is defined when the expression under the square root is non-negative:
\[
x + 5 \geq 0 \quad \Rightarrow \quad x \geq -5
\]
Since \( f(x) \) is a polynomial, it is defined for all real numbers.
Therefore, the domain of \( (f \circ g)(x) \) is determined by the domain of \( g(x) \), which is:
\[
[-5, \infty)
\]
Now we can summarize our result:
\[
(f \circ g)(x) = x + 7; \quad \text{domain: } [-5, \infty)
\]
Thus, the correct response is:
**\( (f \circ g)(x) = x + 7; \quad \text{domain: } [-5, \infty) \)**
We have the following functions:
- \( f(x) = x^2 + 2 \)
- \( g(x) = \sqrt{x + 5} \)
Now, we compute \( (f \circ g)(x) \):
\[
g(x) = \sqrt{x + 5}
\]
Substituting \( g(x) \) into \( f(x) \):
\[
f(g(x)) = f(\sqrt{x + 5}) = (\sqrt{x + 5})^2 + 2
\]
Calculating this gives:
\[
(\sqrt{x + 5})^2 = x + 5
\]
Thus, we have:
\[
f(g(x)) = x + 5 + 2 = x + 7
\]
Now we need to determine the domain of \( (f \circ g)(x) \). The domain of \( g(x) \) must be considered first. The function \( g(x) = \sqrt{x + 5} \) is defined when the expression under the square root is non-negative:
\[
x + 5 \geq 0 \quad \Rightarrow \quad x \geq -5
\]
Since \( f(x) \) is a polynomial, it is defined for all real numbers.
Therefore, the domain of \( (f \circ g)(x) \) is determined by the domain of \( g(x) \), which is:
\[
[-5, \infty)
\]
Now we can summarize our result:
\[
(f \circ g)(x) = x + 7; \quad \text{domain: } [-5, \infty)
\]
Thus, the correct response is:
**\( (f \circ g)(x) = x + 7; \quad \text{domain: } [-5, \infty) \)**