Numerical Differentiation: Difference approximation on numerical data

[Daria_Imparo]

Homework Statement

I am given a table of data derived from experiment. A force (F) is applied to a spring and the extension (x) is measured and recorded. An additional column of data for the derivative (dF/dx) is also provided.

Here is the data:
x(m) F(kN) df/dx (kN/m)
0.0 0.0 5.0
0.03 0.3 10.2
0.06 0.5 4.3
0.12 0.7 5.2
0.2 1.8 8.2
0.22 1.9 1.1


The task is: using a difference approximation ( I think this is either backward, forward or central difference approximation), estimate the numerical value of the derivative dF/dx at x=0.16 based on the values provided in the table.

Homework Equations

None given

The Attempt at a Solution

None so far. Since I don't know how to go about it.

I know the solution to this must be very simple, but it just won't filter into my brain.

 

[HallsofIvy]

If you "this is either backward, forward or central difference approximation" you surely must know what those things are! Try them.

 

[Daria_Imparo]

If you "this is either backward, forward or central difference approximation" you surely must know what those things are! Try them.

OK, something has filtered: since x=0.16 is not part of the given data ( and I don't have its corresponding y value), I cannot include x=0.16 as part of the backward,forward or central difference methods.

If I work with the numbers around the 0.16 value, ie. 0.12 and 0.22:
Backward difference: slope = (0.7-0.5)/(0.12-0.06) = 3.33

Forward difference: slope = (1.9-1.8)/(0.22-0.20) = 5

Central Difference therefore: slope = (3.33+5)/2 = 4.17

But this approach cannot be correct, because the value of dF/dx between x=0.12 and x=0.2 is in the range of 5.2 and 8.2. So I would expect dF/dx for x=0.16 to be between 5.2 and 8.2. However, 4.17 obtained is not within that range.

Apparently the data is not necessarily accurate, so maybe this is correct? It will be correct if my approach is correct. And that is the question right now. Can I what I did?