Function is y=x2 .

Take three points, x=2, x=3, x=4.

Approximate this function at these three points for a deviation Δx =0.1. Which of the three points does the approximation works best? Which point does it works worst?

To approximate the function y=x^2 at the three given points, x=2, x=3, and x=4, with a deviation Δx=0.1, we can follow these steps:

1. Calculate the y-values for each of the three points using the original function y=x^2:
- For x=2, y=2^2=4.
- For x=3, y=3^2=9.
- For x=4, y=4^2=16.

2. Determine the neighboring points for each original point with the given deviation Δx=0.1:
- For x=2, the neighboring points are x=2+0.1=2.1 and x=2-0.1=1.9.
- For x=3, the neighboring points are x=3+0.1=3.1 and x=3-0.1=2.9.
- For x=4, the neighboring points are x=4+0.1=4.1 and x=4-0.1=3.9.

3. Calculate the y-values for the neighboring points using the original function y=x^2:
- For x=2.1, y=2.1^2=4.41.
- For x=1.9, y=1.9^2=3.61.
- For x=3.1, y=3.1^2=9.61.
- For x=2.9, y=2.9^2=8.41.
- For x=4.1, y=4.1^2=16.81.
- For x=3.9, y=3.9^2=15.21.

4. Calculate the deviations between the neighboring points' y-values and the original points' y-values.
- For x=2, the deviation between y=4 and the neighboring points' y-values is:
- Deviation with x=2.1: Δy=4.41-4=0.41.
- Deviation with x=1.9: Δy=3.61-4=-0.39.
- For x=3, the deviation between y=9 and the neighboring points' y-values is:
- Deviation with x=3.1: Δy=9.61-9=0.61.
- Deviation with x=2.9: Δy=8.41-9=-0.59.
- For x=4, the deviation between y=16 and the neighboring points' y-values is:
- Deviation with x=4.1: Δy=16.81-16=0.81.
- Deviation with x=3.9: Δy=15.21-16=-0.79.

5. Analyze the deviations to determine which point the approximation works best and worst:
- The point with the smallest deviation (0.39) is x=2 with the neighboring point x=1.9.
- The point with the largest deviation (0.81) is x=4 with the neighboring point x=4.1.

Therefore, the approximation works best at x=2 and worst at x=4.