# DETERMINING THE DOMAIN AND RANGE OF A FUNCTION

Key Ideas :

The domain of a function is the set of values of the independent variable for which the function is defined. The range of a function depends on the equation of the function. The graph depends on the domain and range.

The domain and range of a function can be determined from its graph, from a table of values, or from the function equation. They are usually easier to determine from a graph or a table of values.

Need to Know :

All linear functions include all the real numbers in their domains. Linear functions of the form f(x) = mx + b where ≠ 0, have range {y ∊ R}. Constant functions f(x) = b have range {b}.

• All quadratic functions have domain ∊ R. The range of a quadratic function depends on the maximum or minimum value and the direction of opening.

The domains of square root functions are restricted because the square root of a negative number is not a real number. The ranges are restricted because the square root sign refers to the positive square root. For example,

(i) The function f(x) = √x has domain = {∊ R | x ≥ 0} and range {y ∊ R | y ≥ 0}.

(ii) The function g(x) = √(x - 3) has domain = {∊ R | x ≥ 3} and range {y ∊ R | y ≥ 0}.

When working with functions that model real-world situations, consider whether there are any restrictions on the variables. For example, negative values often have no meaning in a real context, so the domain or range must be restricted to nonnegative values.

Example 1 :

f(x) = 5x + 9

Solution :

This is a linear function, so x and y can be any value.

Domain = {x ∊ R}

Range = {y ∊ R}

Example 2 :

g(x) = -3(x + 1)2 + 6

Solution :

This is a quadratic equation in vertex form. The function has a maximum value at the vertex (-1, 6), x can be any value.

Domain = {x ∊ R}

Range = {y ∊ R | y ≤ 6}

Example 3 :

h(x) = √(2 - x)

Solution :

We cannot take the square root of a negative number, so (2 - x) must be positive or zero.

2 - x ≥ 0

2

Domain = {x ∊ R | x ≤ 2}

√(2 - xmeans the positive square root, so y is never negative.

Range = {y ∊ R | y  0}

Example 4 :

p(x) = 1/(x - 2)

Solution :

The given function is a rational function.

To find the domain of a rational function, we have to find the value of x that makes the denominator zero.

In 1/(x - 2), if we substitute x = 2, the denominator becomes zero and it is undefined.

So, p(x) is defined for all real values of x except x = 2.

Domain of p(x) = R - {0}

To find range of the rational function above, find the inverse of p(x).

p(x) = 1/(x - 2)

y = 1/(x - 2)

Interchange the variables.

x = 1/(y - 2)

Solve for y in terms of x.

(y - 2)x = 1

y - 2 = 1/x

y = 1/x + 2

y = (1 + 2x)/x

y = (2x + 1)/x

p-1(x) = (2x + 1)/x

Find the domain of p-1(x).

In (2x + 1)/x ,if we substitute x = 0, the denominator becomes zero and it is undefined.

So, p-1(x) is defined for all real values of x except x = 0.

Domain of p-1(x) = R - {0}

Range of p(x) = Domain of p-1(x)

Range of p(x) = R - {0}

Example 5 :

A gull landing on the guardrail causes a pebble to fall off the edge. The speed of the pebble as it falls to the ground is a function is v(d) = √(2gd) where d is the distance, in meters, the pebble has fallen, v(d) is the speed of the pebble, in meters per second (m/s) and g is the acceleration due to gravity—about 9.8 meters per second squared (m/s2). Determine the domain and range of v(d), the pebble’s speed.

Solution :

d = 0 when the pebble begins to fall, and d = 346 when it lands.

So, the domain is ≤ d ≤ 346.

The pebble starts with speed 0 m/s.

v(0) = v(d) = √(2 ⋅ 9.8 ⋅ 0) = 0

When the pebble lands, d = 346.

v(346) = √(2 ⋅ 9.8 ⋅ 346)

= √6781.6

82.4

The domain is

{d ∊ R | 0 ≤ d ≤ 346}

and the range is

{v(d) ∊ R | 0 ≤ v(d) ≤ 82.4}

Example 6 :

Vitaly and Sherry have 24 m of fencing to enclose a rectangular garden at the back of their house.

a) Express the area of the garden as a function of its width.

b) Determine the domain and range of the area function.

Solution :

They need fencing on only three sides of the garden because the house forms the last side.

Let the width of the garden be x m.

Then the length is (24 - 2x).

Let A(x) be the area of the garden.

A(x) = x(24 - 2x)

= 24x - 2x2

= -2x2 + 24x

= -2(x2 - 12x)

= -2[x2 - 2(x)(6) + 62 - 62]

= -2[(x - 6)2 - 36]

A(x) = -2(x - 6)2 - 72

The smallest the width can approach is 0 m. The largest the width can approach is 12 m.

Domain = {x ∊ R | 0 < x < 12}

A(x) = -2(x - 6)2 - 72 is a quadratic equation in vertex form and it has a maximum value at the vertex (6, 72).

Range = {A(x) ∊ R | 0 < A(x) ≤ 72}

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