Unique $f\left( x \right)$

$f(x)=7$
But For $x>7$ ,
$f(x-1)+f(x+1)=\sqrt { 2 } f(x).................(1)$

Multiplying Equation 1 With $\sqrt { 2 }$ ,
$\sqrt { 2 } f(x-1)+\sqrt { 2 } f(x+1)=2f(x)........(2)$ .

Replacing $x$ by $x-1$ in Equation 1,
$f(x-2)+f(x)=\sqrt { 2 } f(x-1)..................(3)$

Replacing $x$ by $x+1$ in Equation 1,
$f(x+2)+f(x)=\sqrt { 2 } f(x+1).................(4)$

Adding Equations (2),(3) and, we Get
$f(x-2)+f(x+2)=0...............................(5)$

Replacing $x$ by $x+4$ in (5), we get
$f(x+2)+f(x+6)=0................................(6)$

Therefore, equating equation (5) and (6),

$f(x-2)=f(x+6)$ and therefore $f(x)=f(x+8)$

So $f(7)=f(15)=f(23)=f(87)$ Hence $f(87)=\boxed { 7 }$ .

by equation (2) we get f(6) = 6 and f(7) = 7. Substitution f(6) and f(7) to equation (1) to get f(8). Based on equation (1), compute f(9), f(10), and so on. we will have an unique pattern f(7) = f(15) = f(23) = f (31) = ... = f(87) = 7

Manipulate to find the period of function to be T=8

f(87)=f(7)=7 (from equation(2))