Discussion
The first derivative of a function \(f \in C^{\infty}(-3, 3)\) is approximated by an interpolating polynomial of degree 2, using the data
(-1, f(-1)), (0, f(0)) and (2, f(2)).
It is found that
\[ f'(0) \approx -\frac{2}{3}f(-1) + \alpha f(0) + \beta f(2). \]
Then, the value of \(\frac{1}{\alpha\beta}\) is
Answer: D
The value of \(\frac{1}{\alpha\beta}\) is 12.
Differentiate the Lagrange interpolating polynomial on the nodes -1, 0, 2 and evaluate at 0. Each coefficient is the derivative at 0 of the corresponding basis polynomial:
\[ \begin{aligned} \ell_{-1}(x) &= \frac{x(x-2)}{3},\ \ \ell_{-1}'(0) = -\frac{2}{3} \\ \ell_{0}(x) &= -\frac{(x+1)(x-2)}{2},\ \ \ell_{0}'(0) = \frac{1}{2} \\ \ell_{2}(x) &= \frac{x(x+1)}{6},\ \ \ell_{2}'(0) = \frac{1}{6} \end{aligned} \]
The first coefficient matches the given -2/3, which confirms the set-up. So \(\alpha = 1/2\) and \(\beta = 1/6\), giving \(\alpha\beta = 1/12\) and \(1/(\alpha\beta) = 12\).
Check: the coefficients must sum to 0 (a constant has zero derivative): -2/3 + 1/2 + 1/6 = 0. The options 3, 6 and 9 come from reading 1/\(\alpha\) or 1/\(\beta\) alone or from a slip in one basis polynomial.
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