Let's **analyze the problem step by step**: ### Given Information - The function \( F \) is defined for integers \( n \) in \( S = \{-9, ..., 9\} \). - The function is **symmetric**: \( F(-n) = F(n) \). - Some values are given: - \( F(3) = 9 \) - \( F(4) = 5 \) - \( F(5) = 3 \) - \( F(6) = 3 \) - \( F(8) = -2 \) - \( F(9) = -7 \) - **Find:** \( F(0), F(1), F(2), F(7) \). --- ## Step 1: Organize the Known Values Let's list known values for positive \( n \): | n | F(n) | |---|------| | 3 | 9 | | 4 | 5 | | 5 | 3 | | 6 | 3 | | 7 | ? | | 8 | -2 | | 9 | -7 | Because the function is symmetric, we only need to figure out the values for \( n \ge 0 \). --- ## Step 2: Look for Patterns Let's look at the sequence for \( n=3 \) to \( n=9 \): - \( F(3) = 9 \) - \( F(4) = 5 \) - \( F(5) = 3 \) - \( F(6) = 3 \) - \( F(7) = ? \) - \( F(8) = -2 \) - \( F(9) = -7 \) Let's compute the **differences**: - \( F(4)-F(3) = 5-9 = -4 \) - \( F(5)-F(4) = 3-5 = -2 \) - \( F(6)-F(5) = 3-3 = 0 \) - \( F(8)-F(6) = -2-3 = -5 \) (notice we skip F(7)) - \( F(9)-F(8) = -7-(-2) = -5 \) So, the function **decreases** as \( n \) increases, but not at a constant rate. --- ## Step 3: Check for a Quadratic Pattern Suppose the function is quadratic: \( F(n) = an^2 + bn + c \). Let's try to fit a quadratic to three known points. ### Use (3, 9), (4, 5), (5, 3): Set up equations: 1. \( F(3) = a\cdot9 + b\cdot3 + c = 9 \) 2. \( F(4) = a\cdot16 + b\cdot4 + c = 5 \) 3. \( F(5) = a\cdot25 + b\cdot5 + c = 3 \) Write equations: 1. \( 9a + 3b + c = 9 \) 2. \( 16a + 4b + c = 5 \) 3. \( 25a + 5b + c = 3 \) Subtract (1) from (2): \[ (16a + 4b + c) - (9a + 3b + c) = 5 - 9 \\ 7a + b = -4 \quad ...(A) \] Subtract (2) from (3): \[ (25a + 5b + c) - (16a + 4b + c) = 3 - 5 \\ 9a + b = -2 \quad ...(B) \] Subtract (A) from (B): \[ (9a + b) - (7a + b) = (-2) - (-4) \\ 2a = 2 \implies a = 1 \] Plug \( a = 1 \) into (A): \[ 7a + b = -4 \implies 7(1) + b = -4 \implies b = -11 \] Now, plug \( a \) and \( b \) into equation 1: \[ 9a + 3b + c = 9 \implies 9(1) + 3(-11) + c = 9 \\ 9 - 33 + c = 9 \implies c = 9 + 24 = 33 \] So, the **quadratic function is**: \[ F(n) = n^2 - 11n + 33 \] --- ## Step 4: Test the Function at Other Points Test for \( n = 6 \): \[ F(6) = 36 - 66 + 33 = 36 - 66 + 33 = (36 + 33) - 66 = 69 - 66 = 3 \] Matches \( F(6) = 3 \) (given). Test for \( n = 8 \): \[ F(8) = 64 - 88 + 33 = (64 + 33) - 88 = 97 - 88 = 9 \] But the given value is \( F(8) = -2 \). **Mismatch!** Test for \( n = 9 \): \[ F(9) = 81 - 99 + 33 = (81 + 33) - 99 = 114 - 99 = 15 \] But given \( F(9) = -7 \). **Mismatch!** So, the quadratic fit works for \( n = 3,4,5,6 \) but **not for \( n = 8,9 \)**. --- ## Step 5: Consider Piecewise or Broken Pattern Maybe the function **changes its rule** after a certain value. Let's try to fit a **linear function** for \( n = 6,8,9 \): Let \( F(n) = an + b \). Using \( n=6, F(6)=3 \); \( n=8, F(8) = -2 \); \( n=9, F(9) = -7 \): From 6 to 8: \( F \) decreases by 5 (\( 3 \to -2 \)), so per step: \( -2.5 \). From 8 to 9: \( -2 \to -7 \), decrease by 5. So, for \( n \ge 6 \), seems \( F(n) = 3 - 2.5 \cdot (n-6) \): - \( n = 6 \): \( 3 - 2.5 \cdot 0 = 3 \) - \( n = 7 \): \( 3 - 2.5 \cdot 1 = 0.5 \) - \( n = 8 \): \( 3 - 2.5 \cdot 2 = 3 - 5 = -2 \) - \( n = 9 \): \( 3 - 2.5 \cdot 3 = 3 - 7.5 = -4.5 \) (But given is -7) So, **the step size is not constant**. Try another approach. --- ## Step 6: Try Cubic Fit for \( n = 6,8,9 \): Three points: (6,3), (8,-2), (9,-7) Let \( F(n) = an^3 + bn^2 + cn + d \) Plug in values: For \( n=6, F(6)=3 \): \[ 216a + 36b + 6c + d = 3 \] For \( n=8, F(8) = -2 \): \[ 512a + 64b + 8c + d = -2 \] For \( n=9, F(9) = -7 \): \[ 729a + 81b + 9c + d = -7 \] But this is getting complicated. Let's try another approach. --- ## Step 7: Consider Symmetry and Value at \( n=0 \) Given symmetry, \( F(0) \) must be unique. --- ## Step 8: Try to Fit Values for \( n = 3, 4, 5, 6 \) From earlier, the quadratic fit works for \( n=3,4,5,6 \): \( F(n) = n^2 - 11n + 33 \) So, - \( F(3) = 9 - 33 + 33 = 9 \) - \( F(4) = 16 - 44 + 33 = 5 \) - \( F(5) = 25 - 55 + 33 = 3 \) - \( F(6) = 36 - 66 + 33 = 3 \) So, perhaps for \( n \leq 6 \), quadratic; for \( n > 6 \), something else. Let's **interpolate** for \( n=7 \): \( F(