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gomer69

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#1 gomer69
Member since 2003 • 1254 Posts

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
lol k
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gomer69

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#3 gomer69
Member since 2003 • 1254 Posts

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

[f(a+h) - f(a)]/h if the function is f(x) = 2x, then the derivative is 2 using that formula just substitute [f(a+h) - f(a)]/h (2(x+h) - 2x)/h = 2x/h + 2h/h - 2x/h = 2h/h =2

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

 

umm that is wrong tho, that is the right answer for f(x)=2x, if you read further u realize that the the aswer is really 0.693 or ln2 lol...
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#4 gomer69
Member since 2003 • 1254 Posts
i am guessing ure teacher is looking for .693 rather than ln(2), just substitute in numbers for h going to zero after u factor out 2^x (and make x = zero, so that 2^x = 1). if u keep substituting in numbers that go closer to zero than it the answer will get closer to ln(2) or 0.693.
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#5 gomer69
Member since 2003 • 1254 Posts
[QUOTE="gomer69"][QUOTE="gomer69"]i think the answer is (0.693147)*(2^x)gomer69
[f(a+h) - f(a)]/h = (2^(x+h) - 2^x)/h = (2^(x)*(2^h-1))/h (i just factored out 2^x) so now u have 2^x times (2^h-1)/h (2^h - 1)/h is equal to f '(0) just put h = .1, .01, .001, .00000001 into that and it goes to .693 so the final answer is 2^x (0.693)

btw the real answer is ln(2)2^x

so f '(0) = ln(2)
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#6 gomer69
Member since 2003 • 1254 Posts
[QUOTE="gomer69"]i think the answer is (0.693147)*(2^x)gomer69
[f(a+h) - f(a)]/h = (2^(x+h) - 2^x)/h = (2^(x)*(2^h-1))/h (i just factored out 2^x) so now u have 2^x times (2^h-1)/h (2^h - 1)/h is equal to f '(0) just put h = .1, .01, .001, .00000001 into that and it goes to .693 so the final answer is 2^x (0.693)

btw the real answer is ln(2)2^x
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#7 gomer69
Member since 2003 • 1254 Posts
i think the answer is (0.693147)*(2^x)gomer69
[f(a+h) - f(a)]/h = (2^(x+h) - 2^x)/h = (2^(x)*(2^h-1))/h (i just factored out 2^x) so now u have 2^x times (2^h-1)/h (2^h - 1)/h is equal to f '(0) just put h = .1, .01, .001, .00000001 into that and it goes to .693 so the final answer is 2^x (0.693)
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#8 gomer69
Member since 2003 • 1254 Posts
i think the answer is (0.693147)*(2^x)
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#9 gomer69
Member since 2003 • 1254 Posts
[QUOTE="gomer69"]the answer is 2 lolgomer69
[f(a+h) - f(a)]/h if the function is f(x) = 2x, then the derivative is 2 using that formula just substitute [f(a+h) - f(a)]/h (2(x+h) - 2x)/h = 2x/h + 2h/h - 2x/h = 2h/h =2

**** just noticed that it was 2^x, rather than 2x, makes this a bit tougher, but if i remember correctly you have to take the ln or log of both sides...
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#10 gomer69
Member since 2003 • 1254 Posts
the answer is 2 lolgomer69
[f(a+h) - f(a)]/h if the function is f(x) = 2x, then the derivative is 2 using that formula just substitute [f(a+h) - f(a)]/h (2(x+h) - 2x)/h = 2x/h + 2h/h - 2x/h = 2h/h =2