To find the composition [tex]\((f \circ g)(x)\)[/tex] when [tex]\(f(x) = \frac{2}{x+3}\)[/tex] and [tex]\(g(x) = \frac{1}{2x}\)[/tex], we need to follow these steps:
1. Determine [tex]\(g(x)\)[/tex]:
We are given that [tex]\(g(x) = \frac{1}{2x}\)[/tex].
2. Substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]:
We need to find [tex]\(f(g(x))\)[/tex]. This means we substitute the expression for [tex]\(g(x)\)[/tex] into the function [tex]\(f(x)\)[/tex].
3. Calculate [tex]\(f(g(x))\)[/tex]:
Given [tex]\(f(x) = \frac{2}{x + 3}\)[/tex], substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[
f(g(x)) = f\left(\frac{1}{2x}\right)
\][/tex]
Now replace [tex]\(x\)[/tex] in [tex]\(f(x)\)[/tex] with [tex]\(\frac{1}{2x}\)[/tex]:
[tex]\[
f\left(\frac{1}{2x}\right) = \frac{2}{\left(\frac{1}{2x}\right) + 3}
\][/tex]
4. Simplify the expression:
Simplify the denominator:
[tex]\[
\frac{2}{\frac{1}{2x} + 3}
\][/tex]
Find a common denominator to combine the terms in the denominator:
[tex]\[
\frac{1}{2x} + 3 = \frac{1}{2x} + \frac{3 \cdot 2x}{2x} = \frac{1 + 6x}{2x}
\][/tex]
So,
[tex]\[
f\left(\frac{1}{2x}\right) = \frac{2}{\frac{1 + 6x}{2x}}
\][/tex]
5. Invert the fraction in the denominator:
To divide by a fraction, we multiply by its reciprocal:
[tex]\[
f\left(\frac{1}{2x}\right) = 2 \cdot \frac{2x}{1 + 6x} = \frac{4x}{1 + 6x}
\][/tex]
Therefore, the composition [tex]\((f \circ g)(x)\)[/tex] is given by:
[tex]\[
(f \circ g)(x) = \frac{4x}{1 + 6x}
\][/tex]
This is the final simplified form of the composition of the functions [tex]\(f\)[/tex] and [tex]\(g\)[/tex].
