# How do you find the domain and range of f(x)= sqrt(4-x^2) /( x-3)?

Sep 16, 2015

Domain $\left\{x : \mathbb{R} , - 2 \le x \le 2\right\}$

Range $\left\{y : \mathbb{R} , - \frac{2}{\sqrt{5}} \le y \le 0\right\}$

#### Explanation:

Domain is quite evident that $x \ne 3$ and $x \le + 2$ and $- 2 \le x$ which essentially boils down to $\left\{x : \mathbb{R} , - 2 \le x \le 2\right\}$

For finding the range, consider the domain from which it can be seen that y would never have any positive value and at end points +2, -2 it is 0, hence $y \le 0$ on the upper side.

Next square both sides, so that it is a quadratic equation in x,

${x}^{2} \left({y}^{2} + 1\right) - 6 {y}^{2} x + 9 {y}^{2} - 4 = 0$. Solve it for x using quadratic formula, x= (3y^2 +- sqrt( -5y^2 +4) )/((y^2+1)

For a function Real to Real $5 {y}^{2} \le 4$ or $\pm y \le \frac{2}{\sqrt{5}}$. Since it is already settled that $y \le 0$ on the upper side, reject $y \le \frac{2}{\sqrt{5}}$' Hence it should be $- y \le \frac{2}{\sqrt{5}}$ or $y \ge - \frac{2}{\sqrt{5}}$.

The range of the function would this be$\left\{y : \mathbb{R} , - \frac{2}{\sqrt{5}} \le y \le 0\right\}$