Let f:R→R: `f(x)=x+1` and g:R→R: `g(x)=2x−3`. Find `(f−g)(x)`.

Explanation

Let `(f(x)=cos nx "sin" ((5pi)/(n)).f(x)` be periodic. Then,
` f(x+lambda)=f(x),` where `lambda` is period
or `cos(nx+n lambda)"sin"((5x+5lambda)/(n))=cos nx sin((5x)/(n))`
At `x=0`,
`cos n lambda sin((5 lambda)/(n))=0`
If `cos n lambda=0,` then
`n lambda=r pi +(pi)/(2), r in I`
or `n(3pi)=r pi+(pi)/(2) " " ( :' lambda=3 pi)`
or `3n-r=(1)/(2) " " ` (not possible)
Now, let `sin((5 lambda)/(n))=0. ` Then
`(5lambda)/(n)=p pi " "(p in I)`
or `(5(3pi))/(n)=p pi " " ( :' lambda=3pi)`
or ` n=(15)/(p)`
For `p= +-1,+-3,+-5,+-15,` we have, respectively,
`n= +-15,+-5,+-3,+-1 " "( :'n in I)`