If f(x) is a monic quartic polynomial such that f(-1)=-1, f(2)=-4, f(-3)=9, and f(4)=-16, find f(1).
+9519 You can let \(f(x) = x^4 + ax^3 + bx^2 + cx + d\) be the polynomial. You can then use the given information to find the missing coefficients.
Substituting x = -1, 2, -3, 4 respectively gives \(\begin{cases}1-a+b-c+d = -1\\16 + 8a + 4b + 2c + d = -4\\81 - 27a + 9b - 3c + d = 9\\256 + 64a + 16b + 4c + d = -16\end{cases}\).
Solving gives:
So f(1) = \(1 + \left(-\dfrac{79}{35}\right) + \left(-\dfrac{89}7\right) + \dfrac{432}{35} + \dfrac{768}{35}\) = 711/35.
