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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)=|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 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 (−∞, ∞)
.
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 open paren negative infinity comma infinity close paren .

(f∘g)(x)=2|x+2|+5
and the domain of (f∘g)(x)
is (−∞, ∞)
.

1 answer

To find \((f \circ g)(x)\), we need to substitute \(g(x)\) into \(f(x)\).

Given:

  • \(f(x) = 2x - 1\)
  • \(g(x) = |x + 2| + 3\)

Now, let's compute \((f \circ g)(x)\):

  1. Start with \(g(x)\): \[ g(x) = |x + 2| + 3 \]

  2. Substitute \(g(x)\) into \(f(x)\): \[ (f \circ g)(x) = f(g(x)) = f(|x + 2| + 3) \]

  3. Replace \(x\) in \(f(x)\) with \(g(x)\): \[ f(|x + 2| + 3) = 2(|x + 2| + 3) - 1 \]

  4. Simplify the expression: \[ = 2|x + 2| + 6 - 1 = 2|x + 2| + 5 \]

So, we have: \[ (f \circ g)(x) = 2|x + 2| + 5 \]

Now, we need to determine the domain of \((f \circ g)(x)\):

Since \(g(x) = |x + 2| + 3\) is defined for all \(x\) (the absolute value function is defined for all real numbers), and \(f(x)\) is also defined for all real numbers, the composition \((f \circ g)(x)\) is defined for all \(x\) too.

Thus, the domain of \((f \circ g)(x)\) is: \[ (-\infty, \infty) \]

Final Answer:

\[ (f \circ g)(x) = 2|x + 2| + 5 \text{ and the domain of } (f \circ g)(x) \text{ is } (-\infty, \infty). \]