1. Lagrange

Using the Lagrange method to find the polynomial that interpolates the points:

f(x) | -3 | -2 | -1 | 0 | 1 |
x | -197 | -35 | -1 | 1 | 5 |

I find:
$p(x) = y_0 L_0(x) + y_1 L_1(x) + y_2 L_2(x) + y_3 L_3(x) + y_4 L_4(x)$

$L_0(x) = \frac{x^4 + 30x^3 - 176x^2 -30x -175}{1269952992}$

$L_1(x) = \frac{x^4 + 192x^3 -986x^2 -192x + 985}{-7931520}$

$L_2(x) = \frac{x^4 +228x^3 -402x^2 -4750x + 4925}{779968}$

$L_3(x) = \frac{x^4 +228x^3 + 52x^2 -5100x -4925}{-57024}$

$L_4(x) = \frac{x^4 +232x^3 +984x^2 -232x -985}{193920}$

So I got the following polynomial:
$p(x) = 4.124x10^{-4}x^4 + 9,55x10^{-4}x^3 +5,34x10^{-3}x^2 +4,845x10^{-3}x + 0,0111$

Is this correct what I did?

2. I would say no, because your

$p(-197)=614036\not=-3.$

Can you show all your intermediate steps?

3. Are you sure you don't have "x" and "f(x)" reversed?

4. Originally Posted by HallsofIvy
Are you sure you don't have "x" and "f(x)" reversed?
The problem statement is the same. But I'm also thinking that it is upside down

5. Using Newton's interpolation I find:

$p_4(x) = 3.216x10^{-6}x^4 +8.11522x10^{-4}x^3 + 0,0375x^2 +0,4991x -0,53807$

It seems more reasonable. But the method of Lagrange I could not.

6. You need to show your work. If you don't do that, we can't help you very much. Here's a page on the Lagrange Interpolating Polynomial.

7. Originally Posted by Ackbeet
You need to show your work. If you don't do that, we can't help you very much. Here's a page on the Lagrange Interpolating Polynomial.
Thanks. I managed to solve the problem.

8. Ok, have a good one.