# 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}

### 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 x_{1,2}are real if 36 y

^{2}+ 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 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 = (x + 2) / (x

^{2}- 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 , +∞)

## 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 . Author - e-mail

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