I was solving this question : $I = \int_0^1xf(x)\,dx = \frac{1}{6}$ $J = \int_0^1 (f(x))^2\,dx = \frac{1}{12}$ $f\left( \frac{1}{2} \right) = ?$ Here $f(x)$ is continuous. So, to solve this I tried to get an integral that contained both the given integrals. So, I assume a parameter $t$ and : $\int_0^1 (f(x) - tx)^2\,dx =0 $ $\int_0^1(f(x))^2,dx -2t\int_0^1xf(x)\,dx +t^2\int_0^1x^2\,dx =0$ Putting in the values and solving, I got : $t = \frac{1}{2}$. So then, $(f(x) - 0.5x)^2$ is always positive, so in order for the integral I assumed to evaluate to $0$, the function had to be $0$. $(f(x)-0.5x)^2 =0$ $f(x) = 0.5x$ $f(0.5) = 0.25$ This was correct according to the answer key, but my doubt is that : There can be another function $g(x) \neq tx$ which also satisfies the given conditions, and $g(0.5) \neq 0.25$. So, how can we prove that either $tx$ is the only function satisfying the given conditions, or that for every possible $g(x)$, $g(0.5)$ will have to be 0.25 ?
							
I was solving this question : 
 
$I = \int_0^1xf(x)\,dx = \frac{1}{6}$
 
$J = \int_0^1 (f(x))^2\,dx = \frac{1}{12}$
 
$f\left( \frac{1}{2} \right) = ?$
 
Here $f(x)$ is continuous.
So, to solve this I tried to get an integral that contained both the given integrals. So, I assume a parameter $t$ and  : 
 
$\int_0^1 (f(x) - tx)^2\,dx =0 $
 
$\int_0^1(f(x))^2,dx -2t\int_0^1xf(x)\,dx +t^2\int_0^1x^2\,dx =0$
 
Putting in the values and solving, I got : $t = \frac{1}{2}$.
 
So then, $(f(x) - 0.5x)^2$ is always positive, so in order for the integral I assumed to evaluate to $0$, the function had to be $0$.
 
$(f(x)-0.5x)^2 =0$
 
$f(x) = 0.5x$
 

$f(0.5) = 0.25$
 
This was correct according to the answer key, but my doubt is that :
 
There can be another function $g(x) \neq tx$ which also satisfies the given conditions, and $g(0.5) \neq 0.25$. So, how can we prove that either $tx$ is the only function satisfying the given conditions, or that for every possible $g(x)$, $g(0.5)$ will have to be 0.25 ?