Question

Find f(2)for f(x)=x^-2 by creatiing a liimit problem using the definition of the derivative and finding the limit through a table of values. Show at least three entries in your limit table.

Answer 1

I think you mean f'(2).

The definition of the derivative is:\[f'(x)=\lim_{h \rightarrow 0}\frac{ f(x+h)-f(x) }{ h }\](the limit of the differential quotient).
So to find f'(2), you need to evaluate:\[f'(2)=\lim_{h \rightarrow 0}\frac{ f(2+h)-f(2) }{ h }=\lim_{h \rightarrow 0}\frac{ \frac{ 1 }{ (2+h)^2 }-\frac{ 1 }{ 2^2 } }{ h }=...\]
Because you don't really have to evaluate the limit, yo can create a table like this:

h f'(2) (approx.)
0.1 ....
0.01 ....
0.001 ....

So for the first approximate value of f'(2), you set h=0.1 in the differential quotient:\[\frac{ \frac{ 1 }{ (2.1)^2 }-\frac{ 1 }{ 2^2 } }{ 0.1 }\]

Do the same for h=0.01 and h=0.001.
You will see a pattern, and you should be able to make a guess of what the limit (=f'(2)) really is.

Answer 2

So then do you mean like this
F(2)
0.2
0.02
0.002

Answer 3

No, f(2) is just 1/2²=1/4 (=0.25).
I think, from reading the text of your problem, you have to find f'(2), so the derivative of f for x=2.
They want you to use the limit definition of the derivative. Look up what I what I wrote about that.

When I do all this, and take e.g. h=0.1, then this first approximation of f'(2) is -0.2324.

Now you have to do the same calculation, only with smaller and smaller values for h, to get an idea what the derivative f'(2) really is.