# 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__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. \( (-\infty , 4-2\sqrt{6} ] \cup [ 4+2\sqrt{6} , +\infty) \)

## More References and links

Find domain and range of functions ,Find the range of functions ,

find the domain of a function and mathematics tutorials and problems . Home Page