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To solve this problem, we will break it into two parts:

1. **Find the first and second-order derivatives at \( x = 2 \) using finite differences.**
2. **Fit a polynomial to the given data points.**

### Step 1: Finding the Derivatives at \( x = 2 \)

Given the dataTo solve this problem, we will break it into two parts:

1. **Find the first and second-order derivatives at \( x = 2 \) using finite differences.**
2. **Fit a polynomial to the given data points.**

### Step 1: Finding the Derivatives at \( x = 2 \)

Given the 
\[
x: 1, 2, 3, 4
\]
\[
y = f(x): -1, 0, 3, 8
\]

#### **First-order derivative (Using Central Difference)**
For the first derivative at \( x = 2 \), the central difference formula is:
\[
f'(x) \approx \frac{f(x+h) - f(x-h)}{2h}
\]
Here, \( h = 1 \) (since the difference between successive \( x \) values is 1). Thus, at \( x = 2 \):
\[
f'(2) \approx \frac{f(3) - f(1)}{2 \times 1} = \frac{3 - (-1)}{2} = \frac{4}{2} = 2
\]

#### **Second-order derivative (Using Central Difference)**
The second derivative at \( x = 2 \) can be approximated using:
\[
f''(x) \approx \frac{f(x+h) - 2f(x) + f(x-h)}{h^2}
\]
Again, \( h = 1 \). Therefore, at \( x = 2 \):
\[
f''(2) \approx \frac{f(3) - 2f(2) + f(1)}{1^2} = \frac{3 - 2 \times 0 + (-1)}{1} = \frac{2}{1} = 2
\]

### Step 2: Fitting a Polynomial

Since there are four data points, we can fit a polynomial of degree 3 (cubic polynomial) to exactly pass through all the points. A cubic polynomial has the form:
\[
y = ax^3 + bx^2 + cx + d
\]

Using the given points \((1, -1)\), \((2, 0)\), \((3, 3)\), and \((4, 8)\), we will set up a system of equations to solve for the coefficients \(a\), \(b\), \(c\), and \(d\).

Let’s calculate the coefficients:

1. At \( x = 1 \), \( -1 = a(1)^3 + b(1)^2 + c(1) + d \) simplifies to: \( a + b + c + d = -1 \).
2. At \( x = 2 \), \( 0 = a(2)^3 + b(2)^2 + c(2) + d \) simplifies to: \( 8a + 4b + 2c + d = 0 \).
3. At \( x = 3 \), \( 3 = a(3)^3 + b(3)^2 + c(3) + d \) simplifies to: \( 27a + 9b + 3c + d = 3 \).
4. At \( x = 4 \), \( 8 = a(4)^3 + b(4)^2 + c(4) + d \) simplifies to: \( 64a + 16b + 4c + d = 8 \).

We will now solve this system of equations to find the values of \( a, b, c, \) and \( d \). Let's proceed to calculate these coefficients.The coefficients of the polynomial are:
\[
a = 0, \quad b = 1, \quad c = -2, \quad d = 0
\]

Thus, the polynomial that fits the data points is:
\[
y = x^2 - 2x
\]

### Summary of the Solution
1. The first derivative at \( x = 2 \) is: \( f'(2) = 2 \).
2. The second derivative at \( x = 2 \) is: \( f''(2) = 2 \).
3. The polynomial that fits the given data points is: \( y = x^2 - 2x \).

Would you like further explanation on any of these steps?

Here are some related questions to consider:
1. How would the finite difference method change if the points were not evenly spaced?
2. Can we use a different polynomial (other than cubic) to fit these points?
3. How do we estimate the error in numerical derivatives using finite differences?
4. What are the conditions for the existence of a unique polynomial passing through a given set of points?
5. How would the derivatives change if a different polynomial was used to approximate the data?

**Tip:** When using finite difference methods, the accuracy improves as the step size (\(h\)) decreases, provided the function is smooth.

Find the first and second order derivative at point x=2 of the function tabulated below and also fit a polynomial.

Learn how to find the first and second derivatives at x=2 using finite differences and fit a polynomial to the data points: (1, -1), (2, 0), (3, 3), and (4, 8). This problem covers numerical differentiation and polynomial fitting, suitable for undergraduate students.

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