f(x) = ax4 + bx2 - x + 7 f(2) = 5 Then f(-2) = ?

For a function written as f(x) = E(x) + O(x), where E(x) collects only the even-power terms (including the constant) and O(x) collects only the odd-power terms, replacing x by -x leaves E(x) unchanged (since (-x) raised to an even power equals x raised to that power) but flips the sign of O(x) (since (-x) raised to an odd power is the negative of x raised to that power). So f(x) and f(-x) differ only through the sign of the odd part -- no need to solve for the individual coefficients.

1. Split f(x) = ax4 + bx2 - x + 7 into its even-power part E(x) = ax4 + bx2 + 7 and its odd-degree part O(x) = -x.

2. At x = 2: O(2) = -2, so f(2) = E(2) + O(2) = E(2) - 2. Given f(2) = 5, this gives E(2) = 7.

3. At x = -2, the even part is unchanged: E(-2) = E(2) = 7. The odd part flips sign: O(-2) = -O(2) = 2.

4. Therefore f(-2) = E(-2) + O(-2) = 7 + 2 = 9.

Cross-check by direct substitution into the original expression:

1. f(2) = a(2)4 + b(2)2 - 2 + 7 = 16a + 4b + 5. Since f(2) = 5, 16a + 4b = 0.

2. f(-2) = a(-2)4 + b(-2)2 - (-2) + 7 = 16a + 4b + 9 = 0 + 9 = 9.

Both methods agree: f(-2) = 9.