
Examples with SolutionsExample 1
Find the Range of function f
defined by
f (x) =  3
Solution to Example 1

The given function has a constant value 3 and therefore the range is the set
{3}
Example 2
Find the Range of function f
defined by
f (x) = 4 x + 5
Solution to Example 2

Assuming that the domain of the given function is the set of all real numbers R, so that the variable x takes all values in the interval
(∞ , +∞)
If x takes all values in the interval (∞ , +∞) then 4 x + 5 takes all values in the interval (∞ , +∞) and the range of the given function is given by the interval
(∞ , +∞)
Example 3
Find the Range of function f
defined by
f (x) = x^{ 2} + 5
Solution to Example 3

Assuming that the domain of the given function is R meaning that x takes all values in the interval (∞ , +∞) which means that x^{ 2} is either zero or positive. Hence we can write the following inequality
x^{ 2} ≥ 0
Add 5 to both sides of the inequality to obtain the inequality
x^{ 2} + 5 ≥ = 0 + 5 or f(x) ≥ 5
The range of f (x) = x^{ 2} + 5 is given by the interval: [5 , +∞)
Example 4
Find the Range of function f
defined by
f (x) = 2 x^{ 2} + 4 x  7
Solution to Example 4

We first write the given quadratic function in vertex form by completing the square
f (x) = 2 x^{ 2} + 4x  7 = 2(x^{ 2}  2 x)  7 = 2( (x  1)^{ 2}  1)  7 = 2(x  1)^{ 2}  5

The domain of the given function is R with x taking any value in the interval (∞ , +∞) hence (x  1)^{ 2} is either zero or positive. We start by writing the inequality
(x  1)^{ 2} ≥ 0

Multiply both sides of the inequality by  2 and change the symbol of inequality to obtain
 2 (x  1)^{ 2} ≤ 0

Add  5 to both sides of the inequality to obtain
 2 (x  1)^{ 2}  5 ≤  5 or f(x) ≤  5 and hence the range of function f is given by the interval (∞  5]
Example 5
Find the Range of function f defined by
f (x) =  3 e^{2x + 5} + 2
Solution to Example 5

The domain of the given function is R and therefore x takes all values in the interval (∞ , +∞). The exponent 2x + 5 takes all values in
the interval (∞ , +∞), we can therefore write the following inequality (basic exponential function always positive)
e^{2x + 5} > 0

Multiply both sides of the inequality by 3 and change symbol of inequality to obtain
 3 e^{2x + 5} < 0

Add 2 to both sides of the inequality to obtain
 3 e^{2x + 5} + 2 < 2 or f(x) < 2
The range of the given function is given by the interval (∞ , 2).
Example 6
Find the Range of function f
defined by
f (x) = 3 e^{x 2 + 5} + 2
Solution to Example 6

The domain of the given function is R, we can start by writing the inequality
^{x 2 ≥ 0
}

Add 5 to both sides of the inequality
^{x 2 + 5 ≥ 5
}

The basic exponential function being an increasing function, we can use the above to write the inequality
e^{ x 2 + 5} ≥ e^{ 5}

Multiply both sides of the inequality by 3 and add 2 to both sides to obtain
 3 e^{ x 2 + 5} + 2 ≤ 3 e^{ 5} + 2

Note the the left hand side of the inequality is equal to f(x). Hence
f(x) < 3 e^{ 5} + 2 which means that the range of function f is given by the interval: (∞ , 3 e^{ 5} + 2]
Example 7
Find the Range of function f
defined by
f (x) = (x  1) / (x + 2)
Solution to Example 7

For this rational function, a direct algebraic method similar to those above is not obvious.
Let us first find its inverse, the domain of its inverse which give the range of f.

We first prove that f is a one to one function and then find its inverse. For a function to be a one to one, we need to show that
If f(a) = f(b) then a = b.
(a  1) / (a + 2) = (b  1) / (b + 2)

Cross multiply, expand and simplify
(a  1)*(b + 2) = (b  1)*(a + 2)
a b + 2 a  b  2 = a b + 2b  a  2
3a = 3b , which finally gives a = b and proves that f is a one to one function.

Let us find the inverse of f
y = (x  1) / (x + 2)

Solve for x
x = (2y + 1) / (1  y)

change y into x and x into y and write the inverse function
f^{ 1} (x) = y = (2x + 1) / (1  x)

The range of f is given by the domain of f^{ 1} and is therefore given by the interval
(∞ , 1) U (1 , + ∞)
Example 8
Find the Range of function f
defined by
f (x) = 3 ln (x + 3)  2
Solution to Example 8

The range of ln (x) is given by the interval (∞ , + ∞). Since the graph of ln(x + 3) is the graph of ln (x) shifted 3 units to the left, the range of ln(x + 3) is also given by the interval (∞ , + ∞).
The graph of 3 ln(x + 3) is that of ln (x + 3) reflected on the xaxis, because of the minus sign and expanded vertically by 3 and therefore the range is still given by the interval (∞ , + ∞).
The graph of f is that of 3 ln(x +3) shifted 2 units up and therefore the range is also given by the interval (∞ , + ∞).
Example 9
Find the Range of function f
defined by
f (x) = 1 / (x^{ 2} + 4)
Solution to Example 9

x^{ 2} is a quantity which may be zero or positive and therefore we can write
x^{ 2} ≥ 0

add 4 to both sides of the inequality to obtain
x^{ 2} + 4 ≥ 4

Divide both side of the inequality x^{ 2} + 4 ≥ 4 by the positive quantity 4(x^{ 2} + 4) to obtain
1/4 ≥ 1 / (x^{ 2} + 4)

The right side of the inequality is equal to f(x). Hence the above inequality may be written as
f(x) ≤ 1/4

Because 1 / (x^{ 2} + 4) is always positive and never zero but may be very close to zero as x increases or decreases indefinitely , the range of f is given by the interval (0; 1/4]
Example 10
Find the Range of function f
defined by
f (x) = √(x^{ 2} + 4x + 4)  6
Solution to Example 10

Note that x^{ 2} + 4x + 4 = (x + 2)^{ 2}. Therefore
f (x) = √(x^{ 2} + 4x + 4)  6 = sqrt( (x + 2)^{ 2} )  6 = x + 2  6

The range of x + 2 is given by the interval [0 , +∞). The graph of f is that of x + 2 shifted down by 6 units and therefore the range of is given by the interval [6 , +∞)
Example 11
Find the Range of function f
defined by
f (x) = 2 sin(3x  π) + 1.5
Solution to Example 11

From basic trigonometry we know that the range of values of sine function is [1 , 1]. Hence
1 ≤ sin(3x  π) ≤ 1

Multiply all terms by 2; change symbols of inequalities and add 1.5 to all terms
2 + 1.5 ≥ 2 sin(3x  π) + 1.5 ≥ 2 + 1.5

Simplify and rewrite as
0.5 ≤ f(x) ≤3.5 which gives the range of f as the interval [0.5 , 3.5].
More References and linksfinding the domain of a function and mathematics tutorials and problems.
