Question

What is the domain and range of `ƒ(x)= (5x+15)/ ((x^2) +1)`?

Answer 1

Refer to explanation

Explanation:

The range is the set of real numbers hence `D(f)=R`.

For the range we set `y=f(x)` and we solve with respect to `x`

Hence

`y=(5x+5)/(x^2+1)=>y*(x^2+1)=5x+5=> x^2*(y)-5x+(y-5)=0`

The last equation is a trinomial with respect to x.In order to have a meaning in real numbers its discriminant must be equal or greater than zero.Hence

`(-5)^2-4*y*(y-5)>=0=>-4y^2+20y+25>=0`

The last is always true for the following values of `y`

`-5/2(sqrt2-1)<=y<=5/2(sqrt2+1)`

Hence the range is

`R(f)=[-5/2(sqrt2-1),5/2(sqrt2+1)]`