Use the functions below.

Question

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 (−∞, ∞)
.

Answer 1

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 \((- \infty, \infty)\).