# Find Range of Rational Functions

Find the range of real valued rational functions using different techniques. There are also matched problems with answers at the bottom of the page.

## Examples with Solutions

### Example 1

Find the Range of function $$f$$ defined by $f(x) = \dfrac{x + 1}{2x-2}$

#### Solution to Example 1

Let us first write the given function as an equation as follows
$$y = \dfrac{x + 1}{2x-2}$$
Solve the above equation for x
$$y (2x - 3) = x + 1$$
$$2 y x - 3 y = x + 1$$
$$2 y x - x = 3 y + 1$$
$$x (2 y - 1) = 3 y + 1$$
$$x = \dfrac{3 y + 1}{2 y - 1}$$
The above expression of x in terms of $$y$$ shows that $$x$$ is real for all real values of $$y$$ except $$\dfrac{1}{2}$$ since $$y = \dfrac{1}{2}$$ will make the denominator $$2 y - 1 = 0$$.
Hence the range of $$f$$, which is the set of all possible values of y, is given by
$$(-\infty , \dfrac{1}{2}) \cup (\dfrac{1}{2} , +\infty)$$
See graph below of function f given above and compare range found and that of the graph.

#### Matched Problem 1:

Find the range of function $$f$$ defined by
$$f(x) = \dfrac{-x + 1}{x - 3}$$
Answer at the bottom of the page.

### Example 2

Find the Range of function $$f$$ defined by $f(x) = \dfrac{x + 2}{x^2 - 9}$

#### Solution to Example 2

Write the given function as an equation
$$y = \dfrac{x + 2}{x^2 - 9}$$
Rewrite the above equation for x in standard form and solve using quadratic formulae
$$y x^2 - x - 9 y - 2 = 0$$
Find the discriminant to the above equation
$$\Delta = (-1)^2 - 4 y (-9 y - 2) = 36 y^2 + 8 y + 1$$
Using the quadratic formulas, the above equation gives the solutions
$$x_{1,2} = \dfrac{1\pm \sqrt{36y^2+8y+1}}{2y}$$
The solutions $$x_{1,2}$$ are real if $$36 y^2 + 8 y + 1 \geq 0$$ and $$y \neq 0$$. Hence we need to solve the inequality
$$36 y^2 + 8 y + 1 \geq 0$$
The discriminant of $$36 y^2 + 8 y + 1$$ is equal to
$$8^2 - 4(36)(1) = - 80$$
Since the discriminant is negative, one test value shows that $$36 y^2 + 8 y + 1$$ is always positive. For $$y = 0$$, we need to set $$y= 0$$ in the equation $$y = \dfrac{x + 2}{x^2 - 9}$$ and solve it
$$\dfrac{x + 2}{x^2 - 9} = 0$$
gives solution $$x = -2$$ and therefore $$y = 0$$ is also in the range of $$f$$. Hence the range of $$f$$ is given by
$$(-\infty , +\infty)$$
See graph below of function $$f$$ given above and compare range found and that of the graph.

#### Matched Problem 2:

Find the range of function $$f$$ defined by $f(x) = \dfrac{-x - 2}{x ^2 - 5}$ Answer at the bottom of the page.

### Example 3

Find the Range of function $$f$$ defined by
$$f(x) = \dfrac{x + 2}{x^2 + 1}$$

#### Solution to Example 3

Write the above function as an equation.
$$y = \dfrac{x + 2}{x^2 + 1}$$ Rewrite the above as follows.
$$y x^2 - x + y - 2 = 0$$
Solve for x using the quadratic formulae. $$x_{1,2} = \dfrac{1\pm \sqrt{1-4y^2+8y}}{2y}$$ The above solutions are real if the radicand is not negative and $$y$$ not equal to 0. Hence we need to solve the inequality
$$1 - 4 y^2 + 8y \geq 0$$
The solution set to the above inequality is
$$1 - \sqrt{5} / 2 \leq y \leq 1 + \sqrt{5} / 2$$ with $$y = 0$$ excluded.
But if we set $$y$$ to 0 in the first equation, we obtain
$$0 = \dfrac{x + 2}{x^2 + 1}$$
which gives $$x = - 2$$ and hence $$y = 0$$ is also included in the range of $$f$$ for now. Hence the range of $$f$$ is given by the interval
$$[ 1 - \sqrt{5} / 2 , 1 + \sqrt{5} / 2 ]$$
See graph below of function $$f$$ given above and compare range found and that of the graph.

#### Matched Problem 3:

Find the range of function $$f$$ defined by $$f(x) = \dfrac{-x - 2}{x^2 + 6}$$ Answer at the bottom of the page.

### Example 4

Find the Range of function $$f$$ defined by $f(x) = \dfrac{x^2 + 2}{x^2 + 1}$

#### Solution to Example 4

Write the function as an equation $$y = \dfrac{x^2 + 2}{x^2 + 1}$$ rewrite as a quadratic equation in $$x$$
$$x^2( y - 1 ) = 2 - y$$
Solve for x $$x_{1,2} = \pm \sqrt{\dfrac{2-y}{y-1}}$$
The above solutions are real if the radicand is not negative. Hence we need to solve the inequality
$$\dfrac{2 - y}{y - 1} \geq 0$$
whose solution set is given by
$$1 \lt y \leq 2$$
The range of the given function is given by the interval
$$(1 , 2]$$
See graph below of function $$f$$ given above and compare range found and that of the graph.

#### Matched Problem 4:

Find the range of function $$f$$ defined by $f(x) = \dfrac{x^2 + 5}{2x^2 + 1}$ Answer at the bottom of the page.

### Example 5

Find the Range of function $$f$$ defined by $f(x) = \dfrac{1}{x^2 - 1}$

#### Solution to Example 5

Write the function as an equation
$$y = \dfrac{1}{x^2 - 1}$$
rewrite as a quadratic equation in $$x$$
$$x^2y = y + 1$$
and solve for $$x$$
$$x_{1,2} = \pm \sqrt{\dfrac{y+1}{y}}$$
for the solutions to be real, the radicand must be non negative. Hence
$$\dfrac{y + 1}{y} \geq 0$$
The solution set to the above inequality is
$$(-\infty -1 ] \cup (0 , +\infty)$$
which is also the range of the given function.
See graph below of function $$f$$ given above and compare range found and that of the graph.

#### Matched Problem 5:

Find the range of function $$f$$ defined by $f(x) = \dfrac{-3}{x^2 - 4}$ Answer at the bottom of the page.

### Example 6

Find the Range of function $$f$$ defined by $f(x) = \dfrac{2 x^2 - 1}{x + 1}$

#### Solution to Example 6

Write the function as an equation
$$y = \dfrac{2 x^2 - 1}{x + 1}$$
rewrite the above as a quadratic function
$$2 x^2 - y x - y - 1 = 0$$
solve for x
$$x_{1,2} = \dfrac{y \pm \sqrt{y^2+8y+8} }{2}$$
the above solutions are real if the radicand is not negative. Hence we need to solve the inequality
$$y^2 + 8 y + 8 \geq 0$$
whose solution set is given by the interval
$$(-\infty , - 4 - 2\sqrt{2} ] \cup [ - 4 + 2\sqrt{2} , +\infty)$$
which is also the range of the given function
See graph below of function $$f$$ given above and compare range found and that of the graph.

#### Matched Problem 6:

Find the range of function $$f$$ defined by $f(x) = \dfrac{2 x^2 + 1}{x - 1}$ Answer at the bottom of the page.

## Answers to the Above Matched Problems

1. $$(-\infty , -1) \cup (-1 , +\infty)$$
2. $$(-\infty , +\infty)$$
3. $$[ \dfrac{2 - \sqrt{10}}{-12} , \dfrac{2 + \sqrt{10}}{-12} ]$$
4. $$(1/2 , 5]$$
5. $$(-\infty , 0 ) \cup [ 3/4 , +\infty)$$
6. $$(-\infty , 4-2\sqrt{6} ] \cup [ 4+2\sqrt{6} , +\infty)$$

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