Find f(8) 
X    2     5   9
f(x) 11 29 32
Solve using lagranges interpolation

To find \( f(8) \) using **Lagrange's Interpolation Formula**, we interpolate the given data points using the formula:

\[
f(x) = \sum_{i=1}^{n} f(x_i) L_i(x)
\]
where \( L_i(x) \) is the **Lagrange Basis Polynomial** for the \( i \)-th point, defined as:
\[
L_i(x) = \prod_{\substack{j=1 \\ j \neq i}}^{n} \frac{x - x_j}{x_i - x_j}
\]

### Given data:
\[
x_1 = 2, \, f(x_1) = 11 \\
x_2 = 5, \, f(x_2) = 29 \\
x_3 = 9, \, f(x_3) = 32
\]

We compute \( L_1(x), L_2(x), L_3(x) \):

#### 1. \( L_1(x) \):
\[
L_1(x) = \frac{x - x_2}{x_1 - x_2} \cdot \frac{x - x_3}{x_1 - x_3} = \frac{x - 5}{2 - 5} \cdot \frac{x - 9}{2 - 9}
\]
\[
L_1(x) = \frac{(x - 5)(x - 9)}{(-3)(-7)} = \frac{(x - 5)(x - 9)}{21}
\]

#### 2. \( L_2(x) \):
\[
L_2(x) = \frac{x - x_1}{x_2 - x_1} \cdot \frac{x - x_3}{x_2 - x_3} = \frac{x - 2}{5 - 2} \cdot \frac{x - 9}{5 - 9}
\]
\[
L_2(x) = \frac{(x - 2)(x - 9)}{(3)(-4)} = \frac{-(x - 2)(x - 9)}{12}
\]

#### 3. \( L_3(x) \):
\[
L_3(x) = \frac{x - x_1}{x_3 - x_1} \cdot \frac{x - x_2}{x_3 - x_2} = \frac{x - 2}{9 - 2} \cdot \frac{x - 5}{9 - 5}
\]
\[
L_3(x) = \frac{(x - 2)(x - 5)}{(7)(4)} = \frac{(x - 2)(x - 5)}{28}
\]

### Interpolation Formula:
\[
f(x) = f(x_1)L_1(x) + f(x_2)L_2(x) + f(x_3)L_3(x)
\]
Substitute \( f(x_1) = 11 \), \( f(x_2) = 29 \), and \( f(x_3) = 32 \):

\[
f(x) = 11 \cdot \frac{(x - 5)(x - 9)}{21} + 29 \cdot \frac{-(x - 2)(x - 9)}{12} + 32 \cdot \frac{(x - 2)(x - 5)}{28}
\]

### Find \( f(8) \):
Substitute \( x = 8 \):

#### 1. Compute \( L_1(8) \):
\[
L_1(8) = \frac{(8 - 5)(8 - 9)}{21} = \frac{(3)(-1)}{21} = -\frac{3}{21} = -\frac{1}{7}
\]

#### 2. Compute \( L_2(8) \):
\[
L_2(8) = \frac{(8 - 2)(8 - 9)}{-12} = \frac{(6)(-1)}{-12} = \frac{-6}{-12} = \frac{1}{2}
\]

#### 3. Compute \( L_3(8) \):
\[
L_3(8) = \frac{(8 - 2)(8 - 5)}{28} = \frac{(6)(3)}{28} = \frac{18}{28} = \frac{9}{14}
\]

#### Substitute into \( f(8) \):
\[
f(8) = 11 \cdot \left(-\frac{1}{7}\right) + 29 \cdot \left(\frac{1}{2}\right) + 32 \cdot \left(\frac{9}{14}\right)
\]
\[
f(8) = -\frac{11}{7} + \frac{29}{2} + \frac{288}{14}
\]

Simplify each term:
- \( -\frac{11}{7} = -\frac{22}{14} \)
- \( \frac{29}{2} = \frac{203}{14} \)
- \( \frac{288}{14} = \frac{288}{14} \)

Combine:
\[
f(8) = -\frac{22}{14} + \frac{203}{14} + \frac{288}{14}
\]
\[
f(8) = \frac{-22 + 203 + 288}{14} = \frac{469}{14}
\]

### Final Answer:
\[
f(8) = \frac{469}{14} \approx 33.5
\]

---

Would you like more details or have any questions?  

Here are 5 related questions to explore:
1. How does Lagrange's interpolation compare to Newton's divided difference method?
2. Can we derive \( f(x) \) as a general polynomial expression from these points?
3. What happens if one of the points is removed from the data set?
4. How can we verify the accuracy of the interpolation at \( x = 8 \)?
5. What are the limitations of using Lagrange's interpolation?

**Tip:** When applying Lagrange's formula, simplify \( L_i(x) \) before substituting \( x \) to avoid computational errors.

Find f(8) using Lagrange's interpolation for the data points: X = {2, 5, 9} and f(X) = {11, 29, 32}.

Discover how to use Lagrange's Interpolation Formula to find f(8) for the data points X = {2, 5, 9} and f(X) = {11, 29, 32}. This step-by-step guide demonstrates polynomial approximation concepts, including basis polynomial calculations, and provides a detailed solution.

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