Approximating f '(2) with a Special Calculator

[char808]

Homework Statement

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.

Homework Equations

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

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?

 

[Mark44]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

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

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

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)?

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

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

 

[char808]

Duh. Thank you.

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

 

[Mark44]

Yes, and you can get even closer...

 

[char808]

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.

 

[Mark44]

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