Unit 1

Functions

Domain and Range

The domain of a function refers to the set of possible

x

-values (input values).

The range of a function refers to the set of possible

y

-values or

f (x)

-values (output values).

For the function

f (x) = \sqrt{x}

, consider whether there are any restrictions to the domain or range.

The table of values showed

x

-values of

1

4

9

, and

16

along with the corresponding function values. You may have noticed that all four of those input values are perfect squares. As such, when the function was evaluated for those values, the results were the whole numbers

1, 2, 3

, and

4

.

Now, consider a non-perfect square input value such as

x = 10

\begin{array}{rcl} f (10) & = & \sqrt{10} \\ ≐ & 3.16 \end{array}

Can

x = 0

be used?

\begin{array}{rcl} f (0) & = & \sqrt{0} \\ = & 0 \end{array}

Zero and all positive perfect and non-perfect square input values are permissible, and therefore belong to the domain of the function

f (x) = \sqrt{x}

.

How about

x = - 1

f (- 1) = \sqrt{- 1}

Try it on your calculator.

The result is an error. Why?

Think about what the operation of taking the square root actually means. It is the opposite operation of squaring a number.

Consider

3^{2}

3^{2} = 9

Then, take the square root of

9

\sqrt{9} = 3

Also, consider

{(- 3)}^{2}

{(- 3)}^{2} = (- 3) (- 3) = 9

Squaring a positive value produces a positive result. Squaring a negative value produces a positive result. No real number multiplied by itself will result in a negative product. Therefore, it is not possible to take the square root of a negative number and get a real output.

These findings confirm there must be a restriction on the domain and range of the function. In the following example, you will determine the specific nature of those restrictions.

Example 1

Determine the restrictions on the variable

x

for each of the following functions.

ii.

State the domain and range of each function.

y = \sqrt{x}

y = \sqrt{x - 3}

y = \sqrt{2 x + 6}

f (x) = 3 \sqrt{4 - x} - 12

Solutions

y = \sqrt{x}

Variable restrictions:

The value under the square root sign cannot be negative. Therefore,

x

must be greater than or equal to

0

ii.

Domain:

\{x ∣ x \geq 0, x \in R\}

(which reads:

x

such that

x

is greater than or equal to zero, and

x

is a real number).

Because

y

increases as

x

increases for

y = \sqrt{x}

, using the smallest possible

x

-value will generate the smallest possible

y

-value. Similarly, using the largest possible

x

-value will generate the largest possible

y

-value. Determining these end

y

-values will allow you to determine the range.

In this case, there is no upper limit on the domain, so select another value for

x

that is in the domain, and generate a corresponding

y

-value to determine the direction of the range.

For the smallest possible

x

-value, zero, we already know the corresponding function value, zero.

For another

x

-value in the domain, try

x = 1

\begin{array}{rcl} y & = & \sqrt{1} \\ = & 1 \end{array}

x

increases in value, so does

y

. Now, the range can be determined. As

x

gets larger (approaches infinity),

y

also gets larger and approaches infinity.

Range:

\{y ∣ y \geq 0, y \in R\}

y = \sqrt{x - 3}

Variable restrictions:

The expression

(x - 3)

is under the square root sign.

x - 3 \geq 0

Solve for the values of

x

that will make

x - 3

greater than or equal to

0

. Add

3

to both sides of the inequality sign.

\begin{array}{rcc} x - 3 & \geq & 0 \\ x & \geq & 3 \end{array}

ii.

Domain:

\{x ∣ x \geq 3, x \in R\}

Determine one endpoint of the range.

\begin{array}{rcl} y & = & \sqrt{x - 3} \\ = & \sqrt{(3) - 3} \\ = & \sqrt{0} \\ = & 0 \end{array}

Now, select another domain value to generate a corresponding

y

-value and the direction of the range.

Try

x = 4

\begin{array}{rcl} y & = & \sqrt{(4) - 3} \\ = & \sqrt{1} \\ = & 1 \end{array}

x

increases in value, so does

y

.

Range:

\{y ∣ y \geq 0, y \in R\}

y = \sqrt{(2 x + 6)}

Variable Restrictions:

The expression

(2 x + 6)

is under the square root sign.

\begin{array}{rcl} 2 x + 6 & \geq & 0 \\ 2 x & \geq & - 6 \\ x & \geq & - 3 \end{array}

ii.

Domain:

\{x ∣ x \geq - 3, x \in R\}

\begin{array}{rcl} y & = & \sqrt{2 x + 6} \\ = & \sqrt{2 (- 3) + 6} \\ = & \sqrt{0} \\ = & 0 \end{array}

Now, select another domain value to generate a corresponding

y

-value and the direction of the range.

Try

x = - 1

\begin{array}{rcl} y & = & \sqrt{2 (- 1) + 6} \\ = & \sqrt{4} \\ = & 2 \end{array}

x

increases, so does

y

.

Range:

\{y ∣ y \geq 0, y \in R\}

f (x) = 3 \sqrt{4 - x} - 12

Variable Restrictions:

The expression

(4 - x)

is under the square root sign.

\begin{array}{rcl} 4 - x & \geq & 0 \\ 4 & \geq & x \end{array}

ii.

Domain:

\{x ∣ x \leq 4, x \in R\}

\begin{array}{rcl} f (x) & = & 3 \sqrt{4 - x} - 12 \\ f (4) & = & 3 \sqrt{4 - 4} - 12 \\ = & 3 (0) - 12 \\ = & - 12 \end{array}

Now, select another domain value to generate a corresponding

f (x)

-value and the direction of the range.

Try

x = 0

\begin{array}{rcl} f (x) & = & 3 \sqrt{4 - x} - 12 \\ f (0) & = & 3 \sqrt{4 - x} - 12 \\ = & 3 (2) - 12 \\ = & - 6 \end{array}

x

decreases, the value of the function increases.

Range:

\{f (x) ∣ f (x) \geq - 12, f (x) \in R\}