1. Period

function , is a real number .

1 ) Find the period of this function
2 ) Give the exempl for .

2. Image does not show, happens with a lot of your posts mate!

3. Originally Posted by dhiab
function , is a real number .

1 ) Find the period of this function
2 ) Give the exempl for .

This doesn't make any sense.

4. The correct question is as follows:
If $f(x+a)=\frac{1}{2}+\sqrt{f(x)-(f(x))^2}$
then prove that
$f(x+2a)=f(x)$

Solution:

Replace $x$ by $x+a$ in the above expression

$f(x+a+a)=\frac{1}{2}+\sqrt{f(x+a)-(f(x+a))^2}$

$
f(x+2a)=\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{f(x)-(f(x))^2}-\left(\frac{1}{2}+\sqrt{f(x)-(f(x))^2}\right)^2}
$

$
f(x+2a)=\frac{1}{2}+\sqrt{\frac{1}{4}-f(x)+(f(x))^2}=\frac{1}{2}+\sqrt{\left(f(x)-\frac{1}{2}\right)^2}
$

$
f(x+2a)=\frac{1}{2}+f(x)-\frac{1}{2}
$

$
f(x+2a)=f(x)
$

There are two more questions of this type:

(i)If $f(x-1)+f(x+1)=f(x)$ ,then find one of the periods of $f(x)$.

(ii)If $f(x-1)+f(x+1)=\sqrt{3}f(x)$,then find one of the periods of $f(x)$

5. Originally Posted by VonNemo19
This doesn't make any sense.
Hello : exo :
watch follow the sulution ci over Thank you

Originally Posted by pankaj
The correct question is as follows:
If $f(x+a)=\frac{1}{2}+\sqrt{f(x)-(f(x))^2}$
then prove that
$f(x+2a)=f(x)$

Solution:

Replace $x$ by $x+a$ in the above expression

$f(x+a+a)=\frac{1}{2}+\sqrt{f(x+a)-(f(x+a))^2}$

$
f(x+2a)=\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{f(x)-(f(x))^2}-\left(\frac{1}{2}+\sqrt{f(x)-(f(x))^2}\right)^2}
$

$
f(x+2a)=\frac{1}{2}+\sqrt{\frac{1}{4}-f(x)+(f(x))^2}=\frac{1}{2}+\sqrt{\left(f(x)-\frac{1}{2}\right)^2}
$

$
f(x+2a)=\frac{1}{2}+f(x)-\frac{1}{2}
$

$
f(x+2a)=f(x)
$

There are two more questions of this type:

(i)If $f(x-1)+f(x+1)=f(x)$ ,then find one of the periods of $f(x)$.

(ii)If $f(x-1)+f(x+1)=\sqrt{3}f(x)$,then find one of the periods of $f(x)$
Hello : Thank you for correction