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Mathematics
OpenStudy (anonymous):

Let \(f\) be a function satisfying \(f(x+y) = f(x) + f(y) \forall x,y\) and if \(f(x) = x^2 g(x)\) where \(g(x)\) is a continuous function, then find \(f'(x)\).

OpenStudy (anonymous):

i am going to make a guess, that \[f'(x)=c\] some constant. not sure what that has to do with \[f(x)=x^2g(x)\] but if i recall correctly the only function satisfying the first condition is \[f(x)=cx\]

OpenStudy (anonymous):

\[f'(x)=2xg(x)+x^2g'(x)\]

OpenStudy (anonymous):

@zed that is true for any f, right?

OpenStudy (anonymous):

i don't think there are any functions that satisfy \[f(x+y)=f(x)+f(y)\] other than constant multiples

OpenStudy (anonymous):

Here, these are the options 1. g'(x) 2. g(0) 3. g(0) + g'(x) 3. 0

OpenStudy (anonymous):

multiple choice?

OpenStudy (zarkon):

looks like zero to me

OpenStudy (anonymous):

yeah, but i am clueless

OpenStudy (anonymous):

really? why. i am fairly certain \[f'(x)=c\] a constant

OpenStudy (zarkon):

us the definition of the derivative

OpenStudy (zarkon):

*use

OpenStudy (zarkon):

(f(x+h)-f(x))/h (f(x)+f(h)-f(x))/h =f(h)/h

OpenStudy (zarkon):

\[f(h)=h^2g(h)\]

OpenStudy (zarkon):

\[\frac{h^2g(h)}{h}=hg(h)\] take limit as h goes to zero...use squeeze theorem

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