GAARGH Math help please

Any and all help with this problem will be greatly appreciated (This is a calculus problem).

Definition 2:The derivative of a function f at a number a, denoted by f'(a), is:

f'(a) = limit of [f(a+h) - f(a)]/h as h approaches 0

Problem: Let f(x) = 2^x. Estimate the value of f'(0) by using Definition 2 and taking successively smaller values of h.

This is as far as I got:

f'(a) = limit of [f(a+h) - f(a)]/h as h approaches 0

       = limit of [f(h) - f(0)]/h as h approaches 0

Where do I go from there?! Please help me. :cry:

I don't get it!  

Ha!!! now something that's beyound college math it's not hard thought, do you know how to solve limits?

f'(x) = lim h -> 0 [f(x+h) - f(x)]/h      where f(x) = 2^x

Well to actually solve that you'd need to derivate using natural logaritmns but the point is not to solve it just to find f'(0) so I'd guess it would go like this:

f'(0) = Lim h -> 0 [2^( 0+h)-2^(0)]/h

It says by using succesive smaller values of h so...let's do exactly that, let's start from 1 then work our way to 0 to see what happens:

h = 1 

 f'(0) = [2^( 0+1)-2^(0)]/1 = 1

h = 0.9

 f'(0) = [2^( 0+0.9)-2^(0)]/0.9 = 0.9622

h = 0.5

f'(0) = [2^( 0+0.5)-2^(0)]/0.5 = 0.8284

h = 0.2

f'(0) = [2^( 0+0.2)-2^(0)]/0.2  = 0.7435

h = 0.01

f'(0) = [2^( 0+0.01)-2^(0)]/0.01 =  0.6955

h = 0.0001

f'(0) = [2^( 0+0.001)-2^(0)]/0.001 =  0.6933

So it is now safe to say that f'(0) of 2^x is something like 0.7 

Oh crap! I should have mentioned this is calculus (college course).

This problem is killing me. :cry:

Ha!!! now something that's beyound college math it's not hard thought, do you know how to solve limits?85070537783

Yes. Why? Can you help me?

Oh crap! I should have mentioned this is calculus (college course).

This problem is killing me. :cry:

 

The_Ish

Oh well im only in grade 10.....:P 

Oh crap! I should have mentioned this is calculus (college course).

This problem is killing me. :cry:

 

The_Ish

Yeah for the most part none of us are in college. i asked my brother, but he said he dosent feel like doing calculus right now, hes pretty smart but i guess hes tired or something, sorry

Yeah for the most part none of us are in college. i asked my brother, but he said he dosent feel like doing calculus right now, hes pretty smart but i guess hes tired or something, sorry

kevass007

Oh well thanks for trying. :)

Though I am still lost. :?

Oh well, let's see if anyonre can come up with something.

the answer is 2 lolgomer69

[QUOTE="gomer69"]the answer is 2 lolgomer69

Oh crap! I should have mentioned this is calculus (college course).

This problem is killing me. :cry:

The_Ish

it's not the derivate of 2x is the derivate of 2^x and it's not 2 it's argh I can't remember I'll look it up...

Edit: I remember now: 

 f(x) = 2^x

f'(x) = (2^x) ln(2)

 f'(0) = [2^(0)][Ln (2)] = 1*ln(2) = 0.6931 wich matches what we allready solved before so f'(0) = ln(2) go to sleep now...

[QUOTE="gomer69"]the answer is 2 lolgomer69

[QUOTE="gomer69"]the answer is 2 lolgomer69

i think the answer is (0.693147)*(2^x)gomer69

[QUOTE="gomer69"]i think the answer is (0.693147)*(2^x)gomer69

[QUOTE="gomer69"][QUOTE="gomer69"]i think the answer is (0.693147)*(2^x)gomer69

[QUOTE="gomer69"][QUOTE="gomer69"][QUOTE="gomer69"]i think the answer is (0.693147)*(2^x)gomer69

1+1 = 2... there you go problem solved

[QUOTE="gomer69"]the answer is 2 lolgomer69

Holy freaking crap I just realized you are right. I completely forgot about f(x) = 2^x.

::slaps forehead::

Holy crap it all makes sense.

That said, I feel comepletely stupid for ignoring that declaration, despite it being right there at the beginning. And I thank you greatly for helping me out.

You have no idea how greatly. :oops:  :D

yeah, I got the answer.

it goes from f(h)-f(0)/h to 2^h - 1/h. When I approximate by estimating the value of h, I find that I get closer and closer to .693.

In other words, f'(0) = .693 (or as close to it as possible).

And thanks everyone for all the help. :D :D :D

[QUOTE="gomer69"][QUOTE="gomer69"]the answer is 2 lolThe_Ish

Holy freaking crap I just realized you are right. I completely forgot about f(x) = 2^x.

::slaps forehead::

Holy crap it all makes sense.

That said, I feel comepletely stupid for ignoring that declaration, despite it being right there at the beginning. And I thank you greatly for helping me out.

You have no idea how greatly. :oops:  :D

yeah, I got the answer.

it goes from f(h)-f(0)/h to 2^h - 1/h. When I approximate by estimating the value of h, I find that I get closer and closer to .693.

In other words, f'(0) = .693 (or as close to it as possible).

And thanks everyone for all the help. :D :D :D

The_Ish

[QUOTE="The_Ish"]

[QUOTE="gomer69"][QUOTE="gomer69"]the answer is 2 lolgomer69

Holy freaking crap I just realized you are right. I completely forgot about f(x) = 2^x.

::slaps forehead::

Holy crap it all makes sense.

That said, I feel comepletely stupid for ignoring that declaration, despite it being right there at the beginning. And I thank you greatly for helping me out.

You have no idea how greatly. :oops: :D

Oh. Sorry. :oops: As soon as I saw f(x) = 2x I thought "where did that come from?", then I remembered the declaration and forgot about that mistake you made and got carried away. lol