IF F IS A POLYNOMIAL FUNCTION,such that f:R->R such that f(2x)=f ' (x) f " (x) then f(3)=?
@amistre64 @genius12 @oldrin.bataku @phi
@DLS Before I think of a solution, can you remind me what the notation f: R -> R means again
f is a function from R to R i.e real numbers,you input real numbers and you get real numbers as output,codomain->R and domain->R
@genius12 it means \(f\) is a function whose domain and codomain are the real numbers \(\mathbb{R}\)
Ok the way I thought of it was this. If \(\bf f(x) = e^x \ then \ f(2x) = e^{2x}\) This also means that \(\bf f'(x)=f''(x) = e^x \implies f'(x)f''(x)=f(2x)=(e^x)^2=e^{2x}\) Hence:\[\bf f(3)=e^3 \approx 20.086\] @DLS
answer is 12
wot lol
:|
But it works perfectly fine 0,o
Just because the answer is 12 there doesn't prove this answer wrong. This is still a correct answer and if I was teacher, there is no reason for why I wouldn't take this as a solution. @DLS
hmm.
I was first thinking if we could do something with sin(2x) but now that you told me the answer is 12, that wouldn't work.
i thought sin2x and tan2x too..but wont workk
@genius12 gave an entirely valid solution
@DLS Are you sure it's not 12 lol. Honestly I can't think of anything else that would work.
okay :( thanks!
@DLS If you do discover ANOTHER valid solution, be sure to tell me.
yup
nevermind
that doesnt work >_>..
how u get this answer 12
i will post the solution tomrrow
@genius12 @dan815 and everyone..that solution isn't incorrect but my sincere apologies,I forgot to mention that question says it is a polynomial function,you can still attempt if you want :) I got the solution though,but I honestly think you can crack the problem now.
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