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### 14. Diketahui \( f^{-1}(x) = \frac{x}{x+2} \) dan \( g^{-1}(x) = \frac{2}{3x} \). Tentukan nilai \( (f \circ g)(2) \).

To solve this, we need to find the composition of the functions \( f^{-1}(x) \) and \( g^{-1}(x) \) and evaluate it at \( x = 2 \).

1. **Step 1**: Find \( g^{-1}(2) \).
   \[
   g^{-1}(x) = \frac{2}{3x}
   \]
   Substituting \( x = 2 \):
   \[
   g^{-1}(2) = \frac{2}{3(2)} = \frac{2}{6} = \frac{1}{3}
   \]

2. **Step 2**: Now find \( f^{-1}(g^{-1}(2)) = f^{-1}\left(\frac{1}{3}\right) \).
   \[
   f^{-1}(x) = \frac{x}{x+2}
   \]
   Substituting \( x = \frac{1}{3} \):
   \[
   f^{-1}\left(\frac{1}{3}\right) = \frac{\frac{1}{3}}{\frac{1}{3} + 2} = \frac{\frac{1}{3}}{\frac{7}{3}} = \frac{1}{7}
   \]

Thus, \( (f \circ g)(2) = \frac{1}{7} \).

### 15. Diketahui \( f(x) = 2x - 5 \), \( g^{-1}(x) = \frac{x+3}{4} \), dan \( h^{-1}(x) = 8x \). Tentukan nilai \( (f \circ g \circ h)^{-1}(-7) \).

This question asks us to solve for the inverse of the composition \( (f \circ g \circ h)^{-1}(-7) \).

1. **Step 1**: Find \( h^{-1}(-7) \).
   \[
   h^{-1}(x) = 8x
   \]
   Substituting \( x = -7 \):
   \[
   h^{-1}(-7) = 8(-7) = -56
   \]

2. **Step 2**: Find \( g^{-1}(h^{-1}(-7)) = g^{-1}(-56) \).
   \[
   g^{-1}(x) = \frac{x + 3}{4}
   \]
   Substituting \( x = -56 \):
   \[
   g^{-1}(-56) = \frac{-56 + 3}{4} = \frac{-53}{4} = -\frac{53}{4}
   \]

3. **Step 3**: Find \( f(g^{-1}(h^{-1}(-7))) = f\left(-\frac{53}{4}\right) \).
   \[
   f(x) = 2x - 5
   \]
   Substituting \( x = -\frac{53}{4} \):
   \[
   f\left(-\frac{53}{4}\right) = 2\left(-\frac{53}{4}\right) - 5 = -\frac{106}{4} - 5 = -\frac{106}{4} - \frac{20}{4} = -\frac{126}{4} = -\frac{63}{2}
   \]

Thus, \( (f \circ g \circ h)^{-1}(-7) = -\frac{63}{2} \).

### 16. Diketahui \( f(x-3) = 6x - 11 \) dan \( g(x+1) = \frac{2x - 6}{3x + 4} \). Tentukan komposisi fungsi \( (f \circ g)(2a + 3) \).

To find \( (f \circ g)(2a + 3) \), we need to substitute the expression \( 2a + 3 \) into \( g(x) \), then apply \( f(x-3) \).

1. **Step 1**: Find \( g(2a + 3 + 1) = g(2a + 4) \).
   \[
   g(x+1) = \frac{2x - 6}{3x + 4}
   \]
   Substituting \( x = 2a + 3 \):
   \[
   g(2a + 4) = \frac{2(2a+4) - 6}{3(2a+4) + 4} = \frac{4a + 8 - 6}{6a + 12 + 4} = \frac{4a + 2}{6a + 16}
   \]

2. **Step 2**: Now substitute into \( f(x - 3) \), where \( x = \frac{4a + 2}{6a + 16} \).

Thus, the result is \( f \left( \frac{4a + 2}{6a + 16} - 3 \right) \). To simplify further, we would expand and evaluate.

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Would you like further clarification on these solutions? Here are five related questions to deepen your understanding:

1. How do you find the composition of two functions?
2. What are the key steps to solve for the inverse of a function?
3. Why is simplifying complex fractions important in function composition?
4. What role do inverses play in solving composite function problems?
5. How can transformations of functions (like shifting by constants) affect their evaluation?

**Tip**: When dealing with composite functions, always evaluate the inner functions first and work your way outwards!

Step-by-step solutions to inverse and composite function problems, such as solving (f o g)(2) and (f o g o h)^{-1}(-7). Detailed algebraic solutions for students in grades 10-12.

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