# How do you find the domain and range of g(t) = ( 1 + 8t ) / ( 1 + t^2 )?

Nov 1, 2017

The domain is $t \in \mathbb{R}$
The range is $g \left(t\right) \in \left[- 3.531 , 4.531\right]$

#### Explanation:

As $\forall t \in \mathbb{R} , \left(1 + {t}^{2}\right) > 0$

The domain is $t \in \mathbb{R}$

Let $y = \frac{1 + 8 t}{1 + {t}^{2}}$

$y \left(1 + {t}^{2}\right) = 1 + 8 t$

$y + y {t}^{2} = 1 + 8 t$

$y {t}^{2} - 8 t + y - 1 = 0$

This is a quadratic equation in ${t}^{2}$ and in order for this equation to have solutions, the discriminant $\Delta \ge 0$

$\Delta = {b}^{2} - 4 a c = {\left(- 8\right)}^{2} - \left(4\right) \left(y\right) \left(y - 1\right) \ge 0$

$64 - 4 {y}^{2} + 4 y \ge 0$

${y}^{2} - y - 16 \le 0$

$y = \frac{1 \pm \sqrt{1 + 64}}{2} = \frac{1 \pm \sqrt{65}}{2}$

${y}_{1} = \frac{1 + \sqrt{65}}{2} = 4.531$

${y}_{2} = \frac{1 - \sqrt{65}}{2} = - 3.531$

Therefore,

the range is $g \left(t\right) \in \left[- 3.531 , 4.531\right]$

graph{(1+8x)/(1+x^2) [-12.66, 12.65, -6.33, 6.33]}