Asked by hudson
Use the functions below.
f(x)=2x−1
g(x)=|x+2|+3
What is (f∘g)(x)
? What is the domain of (f∘g)(x)
?
(1 point)
Responses
(f∘g)(x)=2|x+2|+5
and the domain of (f∘g)(x)
is x≥−2
.
open paren f circ g close paren times x is equal to 2 divides x plus 2 divides plus 5 and the domain of open paren f circ g close paren times x is x is greater than or equal to negative 2 .
(f∘g)(x)=|2x+1|+3
and the domain of (f∘g)(x)
is x≥−2
.
open paren f circ g close paren times x is equal to start absolute value 2 x plus 1 end absolute value plus 3 and the domain of open paren f circ g close paren times x is x is greater than or equal to negative 2 .
(f∘g)(x)=2|x+2|+5
and the domain of (f∘g)(x)
is (−∞, ∞)
.
open paren f circ g close paren times x is equal to 2 divides x plus 2 divides plus 5 and the domain of open paren f circ g close paren times x is open paren negative infinity comma infinity close paren .
(f∘g)(x)=|2x+1|+3
and the domain of (f∘g)(x)
is (−∞, ∞)
.
f(x)=2x−1
g(x)=|x+2|+3
What is (f∘g)(x)
? What is the domain of (f∘g)(x)
?
(1 point)
Responses
(f∘g)(x)=2|x+2|+5
and the domain of (f∘g)(x)
is x≥−2
.
open paren f circ g close paren times x is equal to 2 divides x plus 2 divides plus 5 and the domain of open paren f circ g close paren times x is x is greater than or equal to negative 2 .
(f∘g)(x)=|2x+1|+3
and the domain of (f∘g)(x)
is x≥−2
.
open paren f circ g close paren times x is equal to start absolute value 2 x plus 1 end absolute value plus 3 and the domain of open paren f circ g close paren times x is x is greater than or equal to negative 2 .
(f∘g)(x)=2|x+2|+5
and the domain of (f∘g)(x)
is (−∞, ∞)
.
open paren f circ g close paren times x is equal to 2 divides x plus 2 divides plus 5 and the domain of open paren f circ g close paren times x is open paren negative infinity comma infinity close paren .
(f∘g)(x)=|2x+1|+3
and the domain of (f∘g)(x)
is (−∞, ∞)
.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To find \((f \circ g)(x)\), we need to evaluate \(f(g(x))\).
1. **Calculate \(g(x)\)**:
\[
g(x) = |x + 2| + 3
\]
2. **Substitute \(g(x)\) into \(f(x)\)**:
Now we substitute \(g(x)\) into \(f(x)\):
\[
f(x) = 2x - 1
\]
Thus:
\[
(f \circ g)(x) = f(g(x)) = f(|x + 2| + 3)
\]
Substituting into \(f\):
\[
(f \circ g)(x) = 2(|x + 2| + 3) - 1
\]
Simplifying this, we have:
\[
(f \circ g)(x) = 2|x + 2| + 6 - 1
\]
\[
(f \circ g)(x) = 2|x + 2| + 5
\]
Now, let's find the **domain of \((f \circ g)(x)\)**:
- The function \(g(x) = |x + 2| + 3\) is defined for all real numbers \(x\), as the absolute value function is defined everywhere, and adding 3 does not restrict the domain.
- The function \(f(x) = 2x - 1\) is also defined for all real numbers.
Since both \(f\) and \(g\) are defined for all real numbers, \((f \circ g)(x)\) is defined for all \(x\). Therefore, the domain of \((f \circ g)(x)\) is:
\[
(-\infty, \infty)
\]
### Final Answer:
\[
(f \circ g)(x) = 2|x + 2| + 5
\]
*and the domain of \((f \circ g)(x)\) is \((- \infty, \infty)\).*
1. **Calculate \(g(x)\)**:
\[
g(x) = |x + 2| + 3
\]
2. **Substitute \(g(x)\) into \(f(x)\)**:
Now we substitute \(g(x)\) into \(f(x)\):
\[
f(x) = 2x - 1
\]
Thus:
\[
(f \circ g)(x) = f(g(x)) = f(|x + 2| + 3)
\]
Substituting into \(f\):
\[
(f \circ g)(x) = 2(|x + 2| + 3) - 1
\]
Simplifying this, we have:
\[
(f \circ g)(x) = 2|x + 2| + 6 - 1
\]
\[
(f \circ g)(x) = 2|x + 2| + 5
\]
Now, let's find the **domain of \((f \circ g)(x)\)**:
- The function \(g(x) = |x + 2| + 3\) is defined for all real numbers \(x\), as the absolute value function is defined everywhere, and adding 3 does not restrict the domain.
- The function \(f(x) = 2x - 1\) is also defined for all real numbers.
Since both \(f\) and \(g\) are defined for all real numbers, \((f \circ g)(x)\) is defined for all \(x\). Therefore, the domain of \((f \circ g)(x)\) is:
\[
(-\infty, \infty)
\]
### Final Answer:
\[
(f \circ g)(x) = 2|x + 2| + 5
\]
*and the domain of \((f \circ g)(x)\) is \((- \infty, \infty)\).*
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.