Find f, to the power minus 1 , left parenthesis, f, of, 2, right parenthesis, minus, 3, f, of, f, of, minus, 2f 
−1
 (f(2))−3f(f(−2)).
xx	f, of, xf(x)
22	minus, 5−5
00	minus, 6−6
minus, 2−2	00
1111	minus, 2−2
1010	22

To find \( f^{-1}(f(2)) - 3f(f(-2)) \), we need to evaluate each component step by step, starting with \( f(x) \).

Given the function values:
- \( f(2) = 22 \)
- \( f(0) = 0 \)
- \( f(-2) = -6 \)
- \( f(1) = -2 \)
- \( f(10) = 22 \)

### Step 1: Find \( f^{-1}(f(2)) \)
Since \( f(2) = 22 \), we need to find \( f^{-1}(22) \). Observing the provided values, we see:
- \( f(10) = 22 \)

Thus, \( f^{-1}(22) = 10 \).

### Step 2: Calculate \( f(f(-2)) \)
Next, we need to evaluate \( f(f(-2)) \). From the values:
- \( f(-2) = -6 \)

Now we evaluate \( f(-6) \). However, we don’t have \( f(-6) \) listed in the values provided, so we cannot determine this part without additional information. Assuming \( f(-6) \) is defined, we denote it as \( f(-6) = a \).

### Step 3: Substitute into the expression
Now we can rewrite the expression:

\[
f^{-1}(f(2)) - 3f(f(-2)) = 10 - 3f(-6) = 10 - 3a
\]

### Conclusion
Without the value of \( f(-6) \), we cannot simplify further. The final expression is:

\[
10 - 3a
\]

### Summary
1. **Result:** \( f^{-1}(f(2)) - 3f(f(-2)) = 10 - 3f(-6) \)
2. **Known values:** \( f(2) = 22 \), \( f(-2) = -6 \), \( f(10) = 22 \)

Do you have any questions about this solution, or is there additional information about \( f(-6) \)? 

### Additional Questions
1. What is the value of \( f(-6) \)?
2. Are there any more function values provided?
3. Would you like help defining \( f(x) \) if it’s a specific function?
4. How does \( f^{-1}(x) \) relate to the function values given?
5. Do you want to explore the implications of inverse functions in more detail?

### Tip
Always ensure to provide enough function values when dealing with inverses, as missing values can lead to incomplete results.

Find f^{-1}(f(2)) - 3f(f(-2)) given specific function values.

Explore the problem of finding f^{-1}(f(2)) - 3f(f(-2)) using given function values. This solution delves into the evaluation of inverse functions and the necessary function values, suitable for students in Grades 9-11.

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