Question:
The values of \(x\) are \(1, 3, 4\) and those of \(f(x)\) are \(1, 27, 64\). What is the polynomial equation by using the Lagrange interpolation formula?
\(x\) के मान \(1, 3, 4\) हैं और \(f(x)\) के मान क्रमशः \(1, 27, 64\) हैं। लैग्रेंज प्रक्षेप सूत्र का उपयोग करके बहुपद समीकरण क्या है?
Options:
- 8x^2 – 19x + 12
- 8x^2 + 9x + 12
- 8x^2 – 9x + 12
- 8x^2 + 9x + 12
- 8x^2 – 19x + 12
- 8x^2 + 9x + 12
- 8x^2 – 9x + 12
- 8x^2 + 9x + 12
Correct Answer: a)
Explanation:
Given points: \((1, 1), (3, 27), (4, 64)\).
By Lagrange’s interpolation formula:
\[ f(x) = 1 \cdot \frac{(x-3)(x-4)}{(1-3)(1-4)} + 27 \cdot \frac{(x-1)(x-4)}{(3-1)(3-4)} + 64 \cdot \frac{(x-1)(x-3)}{(4-1)(4-3)} \]
\[ f(x) = \frac{x^2 – 7x + 12}{6} – \frac{27(x^2 – 5x + 4)}{2} + \frac{64(x^2 – 4x + 3)}{3} \]
Combining terms over common denominator 6:
\[ f(x) = \frac{(x^2 – 7x + 12) – 81(x^2 – 5x + 4) + 128(x^2 – 4x + 3)}{6} \]
\[ = \frac{48x^2 – 114x + 72}{6} = 8x^2 – 19x + 12 \]
दिए गए बिंदु: \((1, 1), (3, 27), (4, 64)\)।
लैग्रेंज प्रक्षेप सूत्र के अनुसार:
\[ f(x) = 1 \cdot \frac{(x-3)(x-4)}{(1-3)(1-4)} + 27 \cdot \frac{(x-1)(x-4)}{(3-1)(3-4)} + 64 \cdot \frac{(x-1)(x-3)}{(4-1)(4-3)} \]
\[ f(x) = \frac{x^2 – 7x + 12}{6} – \frac{81(x^2 – 5x + 4)}{6} + \frac{128(x^2 – 4x + 3)}{6} \]
\[ = \frac{48x^2 – 114x + 72}{6} = 8x^2 – 19x + 12 \]