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

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