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Candy Bars
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498 Posts
Discussion Starter #1
Given that f is a function with the following properties;
i. f(Xsub1+Xsub2) = f(Xsub1)*f(Xsub2) far all Xsub1 and Xsub2
ii. f(x) = 1 = xg(x) where lim x->0 g(x) =1
Prove that f ' (x) = f(x)
 

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HURL SCOUTS
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10,007 Posts
"WE COULD BUT WE WON'T! It's a spaceship damnit, not a prom limousine. If anyone needs me I'll be in the angry-dome"
 

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Uggh
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5,044 Posts
Given that f is a function with the following properties;
i. f(Xsub1+Xsub2) = f(Xsub1)*f(Xsub2) far all Xsub1 and Xsub2
ii. f(x) = 1 = xg(x) where lim x->0 g(x) =1
Prove that f ' (x) = f(x)
ok so y=x

:tonka: I could figure it out but I just got done doing 2 hours of math my brain hurts
 

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Semper Fidelis
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4,541 Posts
f= the following:

f(x)= I want some candy.
 

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More boost than yours:)
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1,252 Posts
I forgot how to do calculus . . . all my books and notes are sitting right next to me on in the book case, but I'm not digging it out. Sorry. You're on your own, just like I was lol.

Get the solution manual.

Maybe later I'll think about it, but not likely.
 

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hoo dat. wat.
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21,594 Posts
Given that f is a function with the following properties;
i. f(Xsub1+Xsub2) = f(Xsub1)*f(Xsub2) far all Xsub1 and Xsub2
ii. f(x) = 1 = xg(x) where lim x->0 g(x) =1
Prove that f ' (x) = f(x)
This is the last place you want to look for help on the internet. Search for calc proof on the internets.
 

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Pokerob is my B*tch!
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11,515 Posts
that's an Inverse function proof. It's more like late AlgbII which is Pre calc.
 

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Candy Bars
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498 Posts
Discussion Starter #17
actually you prove it using using the definition of a derivative, which is calc 1
i figured it out, at least i think i did,
first i solved for x in f(x)=1+xg(x) for x which is x = (f(x)-1)/g(x) then i took the x's in that equation and replaced them with h's
Then i modified the definition of a derivative
f ' (x) = lim h->0 ((f(x+h)-f(x))/h) => lim h->0 ((f(x)*f(h)-f(x))/h),
Then i replaced h in the denominator with
(f(h)-1)/g(h) making the whole equation look like this

lim h->0 ((f(x)*f(h)-f(x))/(f(h)-1)/g(h)
modify that a bit to become
(f(x)*f(h)-f(x))/(f(x)-1)*g(h)


as h ->0 the top turns to just -f(x)
and the bottom becomes -1
so thats -f(x)/-1 which is just f(x)
That means that f '(x) = f(x)
at least i think i did that right, anyone that knows about calc should tell me if it looks right.
 

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GLFWDA member
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2,674 Posts
actually you prove it using using the definition of a derivative, which is calc 1
i figured it out, at least i think i did,
first i solved for x in f(x)=1+xg(x) for x which is x = (f(x)-1)/g(x) then i took the x's in that equation and replaced them with h's
Then i modified the definition of a derivative
f ' (x) = lim h->0 ((f(x+h)-f(x))/h) => lim h->0 ((f(x)*f(h)-f(x))/h),
Then i replaced h in the denominator with
(f(h)-1)/g(h) making the whole equation look like this

lim h->0 ((f(x)*f(h)-f(x))/(f(h)-1)/g(h)
modify that a bit to become
(f(x)*f(h)-f(x))/(f(x)-1)*g(h)


as h ->0 the top turns to just -f(x)
and the bottom becomes -1
so thats -f(x)/-1 which is just f(x)
That means that f '(x) = f(x)
at least i think i did that right, anyone that knows about calc should tell me if it looks right.
wait, what the hell did you do?
 

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Candy Bars
Joined
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498 Posts
Discussion Starter #19
proved that in the for the function with the properties given that f ' (x) = f(x), the properties are given in the beginning of the post
 
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