lol kyeah, 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
gomer69's forum posts
[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[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
Â
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...[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)
[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
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)
[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...
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
Log in to comment