# Approximations to f ' (2)

1. May 4, 2010

### char808

1. The problem statement, all variables and given/known data

You can use the following special computer to determine values f(x) of a certain function f. Use these values to determine which of the following numbers best approximates f '(2).

There is a calculator provided: http://oregonstate.edu/instruct/mth251/cq/Stage5/Onward/mysteryFunction.html

I know it says it is for a quiz, I'm not taking a quiz.

2. Relevant equations

f'(a) = lim x-->a [f(x) - f(a)] / x-a or lim h--> 0 [f(a+h) - f(a)] / h

3. The attempt at a solution

So, I know f(2) = 6, f(3) = 24 etc.

so f'(2) = lim x-->2 [f(x)-f(2)]/x-2

if I put in 2 for "x" then I get a division by zero problem.

I'm lost as to where to go from here. Help?

Last edited: May 4, 2010
2. May 4, 2010

### Staff: Mentor

Re: Differentials

This won't do you much good because you don't know the formula for f(x).

You can get an approximate value for f'(2) by evaluating (f(3) - f(1))/(3 - 1). You can get a better value by evaluating (f(2.5) - f(1.5))/(2.5 - 1.5).

Can you think of how you might get a value that's even closer to f'(2)?

3. May 4, 2010

### char808

Re: Differentials

Duh. Thank you.

f(2.1)-f(1.9) / 2.1 - 1.9 = 11.005

4. May 4, 2010

### Staff: Mentor

Re: Differentials

Yes, and you can get even closer...

5. May 4, 2010

### char808

Re: Differentials

Ok, so for the sake of my personal enrichment...

I've got the answer for the problem, because I'm trying to find one of a five possible answers, one of which is 11.

If I continue to make the values closer to 2 then I should get a value that is ever closer to 11, right? IE- 2.0001 and 1.9999 etc etc.

6. May 4, 2010

### Staff: Mentor

Re: Differentials

Yes, in theory. In reality you might run up against the ability of the online computer to do arithmetic with sufficient precision, maybe.