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The image contains a question with the function \( f(x) = \frac{2}{x} \) and asks to find and simplify the following:

### (a) \( f\left(\frac{a}{4}\right) \)
To solve:
\[
f\left(\frac{a}{4}\right) = \frac{2}{\frac{a}{4}} = \frac{2 \cdot 4}{a} = \frac{8}{a}
\]

### (b) \( f\left(\frac{22}{a}\right) \)
To solve:
\[
f\left(\frac{22}{a}\right) = \frac{2}{\frac{22}{a}} = \frac{2 \cdot a}{22} = \frac{2a}{22} = \frac{a}{11}
\]

### (c) \( \frac{f(3+h) - f(3)}{h} \)
First, find \( f(3+h) \) and \( f(3) \):
\[
f(3+h) = \frac{2}{3+h}, \quad f(3) = \frac{2}{3}
\]
Now compute:
\[
\frac{f(3+h) - f(3)}{h} = \frac{\frac{2}{3+h} - \frac{2}{3}}{h}
\]
Simplifying the numerator:
\[
\frac{2}{3+h} - \frac{2}{3} = \frac{2(3) - 2(3+h)}{(3+h)(3)} = \frac{6 - (6 + 2h)}{(3+h)(3)} = \frac{-2h}{(3+h)(3)}
\]
Now substitute this into the original expression:
\[
\frac{\frac{-2h}{(3+h)(3)}}{h} = \frac{-2}{(3+h)(3)}
\]

Let me know if you'd like any more details or have any questions! Here are five related questions:

1. What is the general process for simplifying complex fractions?
2. How does one compute the difference quotient \( \frac{f(x+h) - f(x)}{h} \) for a given function?
3. How do you determine the domain of \( f(x) = \frac{2}{x} \)?
4. What does the difference quotient represent in calculus?
5. How does the result of part (c) relate to the derivative of the function?

**Tip:** When simplifying expressions involving fractions, always try to find a common denominator to combine terms effectively!

Let f(x) = 2/x. Find (and simplify as much as possible) (a) f(a/4), (b) f(22/a), (c) (f(3+h) - f(3))/h.

This problem explores simplifying the function f(x) = 2/x through various calculations, including function evaluations and computing the difference quotient. Ideal for high school students (Grades 9-12) looking to deepen their understanding of function manipulation and calculus basics.

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