The range of the function `f(x) = (x^2-9)/(x-3); x!=3` is

Explanation

Let `y=(x^2−9)/(x-3)​`

`y=((x+3)(x−3))/(x−3)​`

`y=x+3`

For , `R−{3`}

`y=3+3`

`y=6`

So, `f(x)=6`

so, Range `=R−{6}`