# 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 = (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 1/2 since y = 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
(-∞ , 1 / 2) ∪ (1 / 2 , +∞)
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
Δ = (-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 x1,2 are real if 36 y2 + 8 y + 1 ≥ 0 and y ≠0. Hence we need to solve the inequality
36 y
2 + 8 y + 1 ≥ 0
The discriminant of 36 y2 + 8 y + 1 is equal to
8
2 - 4(36)(1) = - 80
Since the discriminant is negative, one test value shows that 36 y2 + 8 y + 1 is always positive. For y = 0, we need to set y= 0 in the equation y = (x + 2) / (x2 - 9) and solve it
(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
( -∞ , +∞ )
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 ≥ 0
The solution set to the above inequality is
1 - √5 / 2 ≤ y ≤ 1 + √5 / 2 with y = 0 excluded.
But if we set y to 0 in the first equation, we obtain
0 = (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 - √5 / 2 , 1 + √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
(2 - y) / (y - 1) ≥ 0
whose solution set is given by
1 < y ≤ 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
2 y = 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
> (y + 1) / y ≥ 0
>
The solution set to the above inequality is
> (-∞ -1 ] ∪ (0 , +∞)
>
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 ≥ 0
whose solution set is given by the interval
(-∞ , - 4 - 2√2 ] ∪ [- 4 + 2√2 , +∞)
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. (-∞ , -1) ∪ (-1 , +∞)
2. (-∞ , +∞)
3. [ (2 - √10) /-12 , (2 + √10) /-12 ]
4. (1/2 , 5]
5. (-∞ 0 ) ∪ [ 3/4 , +∞)
6. (-∞ , 4-2√6 ] ∪ [4+2√6 , +∞)