Question

Use the functions below.

f(x)=3⋅4x
g(x)=13⋅4x
What is (fg)(x)
? What is the domain of (fg)(x)
?

(1 point)
Responses

(fg)(x)=9
and the domain of (fg)(x)
is x≥3
.
open paren f over g close paren of x is equal to 9 and the domain of open paren f over g close paren of x is x is greater than or equal to 3 .

(fg)(x)=19
and the domain of (fg)(x)
is x≥3
.
open paren f over g close paren of x is equal to 1 ninth and the domain of open paren f over g close paren of x is x is greater than or equal to 3 .

(fg)(x)=9
and the domain of (fg)(x)
is (−∞,∞)
.
open paren f over g close paren of x is equal to 9 and the domain of open paren f over g close paren of x is open paren negative infinity comma infinity close paren .

(fg)(x)=19
and the domain of (fg)(x)
is (−∞,∞)
.
open paren f over g close paren of x is equal to 1 ninth and the domain of open paren f over g close paren of x is open paren negative infinity comma infinity close paren .

Answer 1

To find \((fg)(x)\), we need to multiply the functions \(f(x)\) and \(g(x)\):

1. Define the functions:
\[
f(x) = 3 \cdot 4^x
\]
\[
g(x) = 13 \cdot 4^x
\]

2. Calculate \((fg)(x)\):
\[
(fg)(x) = f(x) \cdot g(x) = (3 \cdot 4^x) \cdot (13 \cdot 4^x)
\]
\[
= 39 \cdot (4^x \cdot 4^x) = 39 \cdot 4^{2x} = 39 \cdot 16^x
\]

Thus, we can express \((fg)(x)\) as:
\[
(fg)(x) = 39 \cdot 16^x
\]

3. Determine the domain of \((fg)(x)\):
Since both \(f(x)\) and \(g(x)\) are defined for all real numbers (as they involve exponential functions), the domain of \((fg)(x)\) is also all real numbers:

\[
\text{Domain of } (fg)(x) = (-\infty, \infty)
\]

Therefore, you can conclude:
\((fg)(x) = 39 \cdot 16^x\) and the domain of \((fg)(x)\) is \((-∞, ∞)\).

Given your options, it looks like none of them specifically match this conclusion properly.