My body ask me for help, so Could any of you solve this for me so that I can to explain. It would be greatly appreciate. Thanks
this one is a little tricky but not that bad.
$\displaystyle{\lim_{x\to a}}~\dfrac {f(x)-f(a)}{\sqrt{x}-\sqrt{a}}=\displaystyle{\lim_{x\to a}}\dfrac {f(x)-f(a)}{\sqrt{x}-\sqrt{a}}\dfrac {\sqrt{x}+\sqrt{a}} {\sqrt{x}+\sqrt{a}}=$
$\displaystyle{\lim_{x\to a}}~\dfrac {-\sqrt{x} f(a)-\sqrt{a} f(x)+\sqrt{a} f(a)+\sqrt{x} f(x)}{x-a}$
now use L'Hopital's rule
$n(x)=-\sqrt{x} f(a)-\sqrt{a} f(x)+\sqrt{a} f(a)+\sqrt{x} f(x)$
$d(x) = x-a$
$\dfrac {dn}{dx}=-\sqrt{a} f'(x)-\frac{f(a)}{2 \sqrt{x}}+\sqrt{x} f'(x)+\frac{f(x)}{2 \sqrt{x}}$
$\dfrac {dd}{dx}=1$
$\displaystyle{\lim_{x\to a}}~-\sqrt{a} f'(x)-\frac{f(a)}{2 \sqrt{x}}+\sqrt{x} f'(x)+\frac{f(x)}{2 \sqrt{x}}=$
$\displaystyle{\lim_{x\to a}}~\dfrac{f(x)-f(a)}{2\sqrt{x}}+f^\prime(x)\left(\sqrt{x}+\sqrt{a }\right)=2 f^\prime(a) \sqrt{a}$ so
$\displaystyle{\lim_{x\to a}}~\dfrac {f(x)-f(a)}{\sqrt{x}-\sqrt{a}}=2 f^\prime(a) \sqrt{a}$
I wouldn't be shocked if there's a much cleverer way to do this.
As Halls noted the derivative of left side is simply $f(x)$ You really should be able to see that immediately from a first glance.
$\dfrac d {dx} \displaystyle{\int_0^x} f(t)~dt = \dfrac d {dx} \left(f(x)\right)^2$
$f(x) = 2 f(x) f^\prime(x)$
This is of course satisfied if $f(x)=0$ otherwise
$f^\prime(x)=\dfrac 1 2$
$f(x)=\dfrac x 2 + C$
checking we find
$\dfrac {x^2}{4} + Cx \overset{?}{=} \dfrac{x^2}{4} + 2Cx + C^2$
and thus it's clear that $C=0$ so
$f(x)=0 \mbox{ or }f(x)=\dfrac x 2$
Hi,
Romsek's conclusion about your first problem is correct, but I believe his proof is incomplete. The problem is that you need the derivative f' to be 1/2 for all x; I don't think this is obvious. Here's my solution: