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To solve the equation \( 2f = \frac{9 - x^3}{4} \) for the given values of \( x = -2, 0, 3, \) and \( 6 \), we will follow these steps:

1. **Substitute each value of \( x \) into the equation** to solve for \( f \).
2. **Simplify** the equation to find the value of \( f \).

### Step 1: Solve for \( f \) when \( x = -2 \)

Substitute \( x = -2 \) into the equation:
\[
2f = \frac{9 - (-2)^3}{4}
\]
\[
2f = \frac{9 - (-8)}{4} = \frac{9 + 8}{4} = \frac{17}{4}
\]
\[
f = \frac{17}{8}
\]

### Step 2: Solve for \( f \) when \( x = 0 \)

Substitute \( x = 0 \) into the equation:
\[
2f = \frac{9 - 0^3}{4}
\]
\[
2f = \frac{9 - 0}{4} = \frac{9}{4}
\]
\[
f = \frac{9}{8}
\]

### Step 3: Solve for \( f \) when \( x = 3 \)

Substitute \( x = 3 \) into the equation:
\[
2f = \frac{9 - 3^3}{4}
\]
\[
2f = \frac{9 - 27}{4} = \frac{-18}{4} = -\frac{9}{2}
\]
\[
f = -\frac{9}{4}
\]

### Step 4: Solve for \( f \) when \( x = 6 \)

Substitute \( x = 6 \) into the equation:
\[
2f = \frac{9 - 6^3}{4}
\]
\[
2f = \frac{9 - 216}{4} = \frac{-207}{4}
\]
\[
f = -\frac{207}{8}
\]

### Final Results:

- \( x = -2 \), \( f = \frac{17}{8} \)
- \( x = 0 \), \( f = \frac{9}{8} \)
- \( x = 3 \), \( f = -\frac{9}{4} \)
- \( x = 6 \), \( f = -\frac{207}{8} \)

Would you like more details or have any questions? 

Here are 5 relative questions:
1. What happens if \( x \) is a non-integer, such as \( x = 2.5 \)?
2. How would the equation change if the coefficient of \( f \) was 3 instead of 2?
3. Can \( f \) ever be zero? Under what conditions?
4. What is the behavior of \( f(x) \) as \( x \) becomes very large or very small?
5. How does changing the constant 9 in the equation affect the values of \( f \)?

**Tip:** When substituting values into equations, make sure to correctly handle negative signs and powers, as they can significantly change the outcome.

Evaluate the equation 2f = (9 - x^3) / 4 for the values x = -2, 0, 3, and 6.

This problem involves solving the equation 2f = (9 - x^3) / 4 for the values x = -2, 0, 3, and 6. Explore step-by-step solutions, including substitution and simplification, suitable for Grades 8-10.

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