If `f: Rrarr[0,oo)`is a function such that `f(x-1)+f(x+1)=sqrt(3)f(x),` then find its period.

Explanation

`f(x-1) +f(x+1)=sqrt(3) f(x) " (1)" `
Putting `x+2` for x in relation (1), we get
`f(x+1) +f(x+3)=sqrt(3) f(x+2) " (1)" `
From (1) and (2), we get
`f(x-1) +2f(x+1)+f(x+3) =sqrt(3)(f(x)+f(x+2))`
`=sqrt(3)(sqrt(3)f(x+1))`
`=3f(x+1)`
`or f(x-1)+f(x+3)=f(x+1) " (3)" `
Putting `x+2` for x in (3), we get
`f(x+1)+f(x+5)=f(x+3) " (4)" `
Adding (3) and (4), we get `f(x-1)= -f(x+5).`
Now, put `x+1` for x. Then `f(x)= -f(x+6) " (5)" `
Put `x+6` in place of x in (5). Then `f(x+6)= -f(x+12).`
Therefore, from (5) again, `f(x)= -[-f(x+12)]=f(x+12)`.
Hence, the period of `f(x)` is 12.