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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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498 Posts

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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678 Posts

7... or maybe 11...

7 + 11 = 7/11

mmmmmm.... slurpee

7 + 11 = 7/11

mmmmmm.... slurpee

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4,182 Posts

its definately 7

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498 Posts

7/11 i dont think thats the answer but now i want a slurpee

ok so y=x

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)

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

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498 Posts

Someone must know this one!

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6,224 Posts

Do your own homework. oke:Someone must know this one!

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33,035 Posts

Prove that it DOESN'T, BeeYotch!!!

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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1,252 Posts

Get the solution manual.

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

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498 Posts

Joined

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21,594 Posts

This is the last place you want to look for help on the internet. Search for calc proof on the internets.

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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11,515 Posts

that's an Inverse function proof. It's more like late AlgbII which is Pre calc.

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498 Posts

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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2,674 Posts

wait, what the hell did you do?

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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498 Posts

Joined

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2,674 Posts

ummmm...sounds great :sonicjay:

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