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

Aug 23, 2015

Domain:
$x \ne \pm 2$
or
$\left(- \infty , - 2\right)$ and $\left(- 2 , 2\right)$ and $\left(2 , + \infty\right)$

Range:
$\left(- \infty , \frac{3 - \sqrt{5}}{8}\right]$ and $\left[\frac{3 + \sqrt{5}}{8} , + \infty\right)$

#### Explanation:

Domain is easy. This function is defined everywhere except those values of argument $x$ when the denominator ${x}^{2} - 4$ equals to zero. These points are solutions to an equation
${x}^{2} - 4 = 0$
or
${x}^{2} = 4$
Solutions are $x = 2$ and $x = - 2$.
At these points the function has vertical asymptotes.

To address the range, let's first transform the function as follows:
$y = \frac{x - 3}{{x}^{2} - 4} = \frac{\left(x - 2\right) - 1}{\left(x - 2\right) \left(x + 2\right)} = \frac{1}{x + 2} - \frac{1}{{x}^{2} - 4}$

Next step is to graph this function as a sum of two graphs $y = \frac{1}{x + 2}$ and $y = - \frac{1}{{x}^{2} - 4}$. If any of these graphs present a problem or you need some explanation about how to sum two graphs, I suggest to refer to a course of mathematics on Unizor topic Algebra - Graphs.

The resulting graph would look like
$y = \frac{1}{x + 2} - \frac{1}{{x}^{2} - 4}$

graph{1/(x+2)-1/(x^2-4) [-10, 10, -5, 5]}

It's easy to see from this graph that the only segment not covered by values of this function is the one between the maximum of the right part of a graph and minimum of the middle part.

To find these values, let's use the calculus to find the arguments where our function reaches its local maximum and minimum values, that is those values of $x$ when the first derivative of our function equals to zero. The greater value of $x$ would correspond to the lower boundary of the segment of values the function does not take, and the smaller value of $x$ would correspond to the top boundary value of this segment.

Taking the derivative of this function results in'
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{x + 2} ^ 2 + \frac{2 x}{{x}^{2} - 4} ^ 2$

Now we have solve the equation
$- \frac{1}{x + 2} ^ 2 + \frac{2 x}{{x}^{2} - 4} ^ 2 = 0$
or
$- \frac{1}{x + 2} ^ 2 + \frac{2 x}{{\left(x - 2\right)}^{2} {\left(x + 2\right)}^{2}} = 0$
or
$- {\left(x - 2\right)}^{2} / \left({\left(x - 2\right)}^{2} {\left(x + 2\right)}^{2}\right) + \frac{2 x}{{\left(x - 2\right)}^{2} {\left(x + 2\right)}^{2}} = 0$
or
$\frac{- {\left(x - 2\right)}^{2} + 2 x}{{\left(x - 2\right)}^{2} {\left(x + 2\right)}^{2}} = 0$

In this case we have to find where the numerator equals to zero, that to solve
$- {\left(x - 2\right)}^{2} + 2 x = 0$
or
$- {x}^{2} + 6 x - 4 = 0$
or
${x}^{2} - 6 x + 4 = 0$
Solutions of this quadratic equation are
${x}_{1} = 3 + \sqrt{5}$ and ${x}_{2} = 3 - \sqrt{5}$

We have to use the larger value ${x}_{1} = 3 + \sqrt{5}$ in our function to get the lower boundary of a segment where there are no values of our function:

$\frac{1}{{x}_{1} + 2} - \frac{1}{{x}_{1}^{2} - 4} = \frac{1}{3 + \sqrt{5} + 2} - \frac{1}{{\left(3 + \sqrt{5}\right)}^{2} - 4} =$

$= \frac{1}{5 + \sqrt{5}} - \frac{1}{10 + 6 \sqrt{5}} = \frac{5 - \sqrt{5}}{25 - 5} - \frac{6 \sqrt{5} - 10}{180 - 100} =$

$= \frac{3 - \sqrt{5}}{8}$

We have to use the smaller value ${x}_{1} = 3 - \sqrt{5}$ in our function to get the upper boundary of a segment where there are no values of our function:

$\frac{1}{{x}_{2} + 2} - \frac{1}{{x}_{2}^{2} - 4} = \frac{1}{3 - \sqrt{5} + 2} - \frac{1}{{\left(3 - \sqrt{5}\right)}^{2} - 4} =$

$= \frac{1}{5 - \sqrt{5}} + \frac{1}{6 \sqrt{5} - 10} = \frac{5 + \sqrt{5}}{25 - 5} + \frac{6 \sqrt{5} + 10}{180 - 100} =$

$= \frac{3 + \sqrt{5}}{8}$

Therefore, the range of our function is
$\left(- \infty , \frac{3 - \sqrt{5}}{8}\right]$ and $\left[\frac{3 + \sqrt{5}}{8} , + \infty\right)$

Aug 23, 2015

Solve for $x$ in terms of $y$ to find the range of $y$.

$\left(- \infty , \frac{3 - \sqrt{5}}{8}\right] \cup \left[\frac{3 + \sqrt{5}}{8} , \infty\right)$

#### Explanation:

Here's an alternative method to try to find the range using elementary methods.

$y = \frac{x - 3}{{x}^{2} - 4}$

Multiply both sides by $\left({x}^{2} - 4\right)$ to get:

$x - 3 = y \left({x}^{2} - 4\right) = y {x}^{2} - 4 y$

Subtract $x - 3$ from both sides to get:

$y {x}^{2} - x + \left(3 - 4 y\right) = 0$

If $y = 0$ then this simplifies to:

$- x + 3 = 0$, giving $x = 3$

Otherwise, use the quadratic formula to get:

$x = \frac{1 \pm \sqrt{1 - 4 y \left(3 - 4 y\right)}}{2 y}$

$= \frac{1 \pm \sqrt{16 {y}^{2} - 12 y + 1}}{2 y}$

This will have Real solutions when the discriminant is non-negative, that is when:

$16 {y}^{2} - 12 y + 1 \ge 0$

This quadratic in $y$ has zeros given by the quadratic formula:

$y = \frac{12 \pm \sqrt{{12}^{2} - 4 \cdot 16 \cdot 1}}{32} = \frac{12 \pm \sqrt{144 - 64}}{32}$

$= \frac{12 \pm \sqrt{80}}{32} = \frac{12 \pm 4 \sqrt{5}}{32} = \frac{3 \pm \sqrt{5}}{8}$

Hence $16 {y}^{2} - 12 y + 1 \ge 0$ for

$y \in \left(- \infty , \frac{3 - \sqrt{5}}{8}\right] \cup \left[\frac{3 + \sqrt{5}}{8} , \infty\right)$