SOLUTION: Suppose the polynomial f(x) is of degree 3 and satisfies f(3) = 2, f(4) = -4, f(5) = 6, and f(8) = 8. Determine the value of f(0).

Question 1209686: Suppose the polynomial f(x) is of degree 3 and satisfies f(3) = 2, f(4) = -4, f(5) = 6, and f(8) = 8.

Determine the value of f(0).
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Let $f(x)$ be a polynomial of degree 3. We are given the following values:
$f(3) = 2$
$f(4) = -4$
$f(5) = 6$
$f(8) = 8$
We can use the Lagrange interpolation formula to find the polynomial $f(x)$:
\[f(x) = \sum_{i=0}^3 y_i L_i(x)\]
where $y_i$ are the given function values and $L_i(x)$ are the Lagrange basis polynomials.
In our case, the points are $(3, 2), (4, -4), (5, 6), (8, 8)$.
\[L_0(x) = \frac{(x-4)(x-5)(x-8)}{(3-4)(3-5)(3-8)} = \frac{(x-4)(x-5)(x-8)}{(-1)(-2)(-5)} = -\frac{1}{10}(x-4)(x-5)(x-8)\]
\[L_1(x) = \frac{(x-3)(x-5)(x-8)}{(4-3)(4-5)(4-8)} = \frac{(x-3)(x-5)(x-8)}{(1)(-1)(-4)} = \frac{1}{4}(x-3)(x-5)(x-8)\]
\[L_2(x) = \frac{(x-3)(x-4)(x-8)}{(5-3)(5-4)(5-8)} = \frac{(x-3)(x-4)(x-8)}{(2)(1)(-3)} = -\frac{1}{6}(x-3)(x-4)(x-8)\]
\[L_3(x) = \frac{(x-3)(x-4)(x-5)}{(8-3)(8-4)(8-5)} = \frac{(x-3)(x-4)(x-5)}{(5)(4)(3)} = \frac{1}{60}(x-3)(x-4)(x-5)\]
Then,
\[f(x) = 2L_0(x) - 4L_1(x) + 6L_2(x) + 8L_3(x)\]
We want to find $f(0)$:
\[f(0) = 2L_0(0) - 4L_1(0) + 6L_2(0) + 8L_3(0)\]
\[L_0(0) = -\frac{1}{10}(-4)(-5)(-8) = -\frac{1}{10}(-160) = 16\]
\[L_1(0) = \frac{1}{4}(-3)(-5)(-8) = \frac{1}{4}(-120) = -30\]
\[L_2(0) = -\frac{1}{6}(-3)(-4)(-8) = -\frac{1}{6}(-96) = 16\]
\[L_3(0) = \frac{1}{60}(-3)(-4)(-5) = \frac{1}{60}(-60) = -1\]
\[f(0) = 2(16) - 4(-30) + 6(16) + 8(-1) = 32 + 120 + 96 - 8 = 240\]
Final Answer: The final answer is $\boxed{240}$

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