f(x)=x-2/x+2 ;g(x)1/x$$ f\left(x\right)= $$

Given the functions $ f(x) = \frac{x-2}{x+2} $ and $ g(x) = \frac{1}{x} $, it seems like you want to find the composition of these functions, specifically $ f(g(x)) $.

The composition $ f(g(x)) $ means substituting $ g(x) $ into $ f(x) $:

\[
f(g(x)) = f\left(\frac{1}{x}\right)
\]

Substitute $ g(x) = \frac{1}{x} $ into $ f(x) = \frac{x-2}{x+2} $:

\[
f\left(\frac{1}{x}\right) = \frac{\left(\frac{1}{x}\right) - 2}{\left(\frac{1}{x}\right) + 2}
\]

Now, simplify the expression:

\[
f\left(\frac{1}{x}\right) = \frac{\frac{1 - 2x}{x}}{\frac{1 + 2x}{x}}
\]

Since the denominators are the same, they cancel out:

\[
f\left(\frac{1}{x}\right) = \frac{1 - 2x}{1 + 2x}
\]

So, the composition of the functions is:

\[
f(g(x)) = \frac{1 - 2x}{1 + 2x}
\]

Do you have any questions or would you like a more detailed explanation?

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Here are some related questions you might find useful:

1. How do you find the domain of $ f(g(x)) $?
2. Can you determine the range of $ f(g(x)) $?
3. What happens if you try to find $ g(f(x)) $?
4. How do you differentiate $ f(g(x)) $?
5. How would you solve for $ x $ if $ f(g(x)) = 0 $?

**Tip:** When composing functions, always be cautious about the domains of the original functions. The composition is only valid where both functions are defined.

Explore the composition of functions f(x) = (x - 2) / (x + 2) and g(x) = 1 / x. Learn how to find and simplify the composition f(g(x)) = (1 - 2x) / (1 + 2x) step-by-step. Suitable for high school students studying functions and their compositions.

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