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

Refer to explanation

#### Explanation:

The range is the set of real numbers hence $D \left(f\right) = R$.

For the range we set $y = f \left(x\right)$ 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

${\left(- 5\right)}^{2} - 4 \cdot y \cdot \left(y - 5\right) \ge 0 \implies - 4 {y}^{2} + 20 y + 25 \ge 0$

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

$- \frac{5}{2} \left(\sqrt{2} - 1\right) \le y \le \frac{5}{2} \left(\sqrt{2} + 1\right)$

Hence the range is

$R \left(f\right) = \left[- \frac{5}{2} \left(\sqrt{2} - 1\right) , \frac{5}{2} \left(\sqrt{2} + 1\right)\right]$