It is very common in mathematics to represent a function by an equation or a formula. We shall present examples and questions, including their solutions, on functions given by equations. Questions related to domain and range of a function are also included.

The equation \( y = 2 x + 1 \) represent a function that assigns exactly one value for \( y \) for each value of \( x \) in the domain of the function. We say that the variable \( y \) is a function of the variale \( x \).

\( y \) is called the dependent variable and \( x \) is called the independent variable since its value depends on \( x \).

Example

Consider the function represented by the equation \( y = 2 x + 1 \).

Let \( x \) take the values in the domain \( D = \{ -2 , 0 , 1 , 2 \} \).

a) Calculate the range \( R \) of values of \( y \) corresponding to the given \( x \) values in the domain \( D \).

b) Represent the function using a table, a set of ordered pairs , a graph and Venn diagrams .

Solution

a)

Calculate the corresponding values of \( y \) for the given values of \( x \).

For \( x = -2 \) , \( y = 2 (-2) + 1 = -3 \)

For \( x = 0 \) , \( y = 2 (0) + 1 = 1 \)

For \( x = 1 \) , \( y = 2 (1) + 1 = 3 \)

For \( x = 2 \) , \( y = 2 (2) + 1 = 5 \)

\( R = \{ -3 , 1 , 3 , 5 \} \)

b)

__Represent the given function as a table__

\( x \) | \( -2 \) | \( 0 \) | \( 1 \) | \( 2 \) |

\( y \) | \( -3 \) | \( 1 \) | \( 3 \) | \( 5 \) |

\( \{ \; (-2,-3) \; , \; (0,1) \; , \; (1,3) \; , \; (2,5) \} \)

The domain \( D \) of a function is the set of all values of the independent variable \( x \) for which the dependent \( y \) variable is defined.

The range \( R \) of a function is the set of all values of the dependent variable \( y \) corresponding to the values of \( x \) in the domain \( D \).

Example

Find the domain of the functions \( y \) as a function of \( x \) defined by the equations:

a) \( y = 2 - x \)

b) \( 2 y - x = 4 \)

c) \( y = \dfrac{1}{x} \)

d) \( y = \sqrt x \)

Solution

a)

The dependent variable \( y \) of the function given by \( y = 2 - x \) is defined for any value \( x \) in the set of the real numbers. Hence the domain of the given function is the __set of all real numbers__.

b)

We first need to solve for the independent variable \( y \)

Rewrite the given equation with the term in \( y \) on the left as follows: \( \quad 2 y = 4 + x \)

Multiply all terms of the above equation by \( \dfrac{1}{2} \): \( \quad \dfrac{1}{2} 2 y = \dfrac{1}{2} 4 + \dfrac{1}{2} x \)

Simplify to obtain: \( \quad y = 2 + \dfrac{1}{2} x \)

The dependent variable \( y \) of the function given by \( y = y = 2 + \dfrac{1}{2} x \) is defined for any value \( x \) in the set of the real numbers. Hence the domain of the given function is the __ set of all real numbers__.

c)

There is division on the right side of the given equation \( y = \dfrac{1}{x} \) and it is known that a division by zero is not allowed in mathematics.
Hence dependent variable \( y \) of the function given by \( y = \dfrac{1}{x} \) is defined for all value \( x \) in the __set of all real numbers except \( x = 0 \)__.

d)

There is square root on the right side of the given equation \( y = \sqrt x \) and it is known that the square root of a negative number in not real in mathematics.
Hence dependent variable \( y \) of the function given by \( y = \sqrt x \) is defined for all value \( x \) in the __set of all non negative real numbers__.

Example

Find the range of the function \( y = x^2 + 1\) for \( x \) in the domain: \( D = \{ 0 , 1 , 6 \} \)

Solution

The range is the set of \( y \) values for \( x \) in the domain. The values in the range are calculated by substituting \( x \) by its values in the domain.

For \( x = 0 \) , \( \quad y = x^2 + 1 = 0^2 + 1 = 0 + 1 = 1 \)

For \( x = 1 \) , \( \quad y = x^2 + 1 = 1^2 + 1 = 1 + 1 = 2 \)

For \( x = 6 \) , \( \quad y = x^2 + 1 = 6^2 + 1 = 36 + 1 = 37 \)

Range: \( \quad R = \{ 1 , 2,37 \} \)

More advanced topics on domain and range are included.

Example - Testing Equations for Functions

Which of the following equations represent \( y \) as a function of \( x \)?

a) \( x + y = 2 \)

b) \( x + 5 = |y| \)

d) \( \dfrac{y}{x} = - 3 \)

c) \( x + y^2 = 25 \)

Solution

a)

Solve the given equation for \( y \) to obtain: \( y = 2 - x \). The given equation __represents__ \( y \) as a function of \( x \) because for each value of \( x \), we have one value of \( y \) only.

b)

Given equation \( | y | = x + 5 \). For both values of \( y = 2 \) and \( y = - 2 \), we obtain \( | y | = 2 \), hence \( x = 5 - |y| = 5 - 2 = 3 \).
Therefore for one input \( x = 3 \), we have two outputs \( y = - 2 \) and \( y = 2 \) and therefore the given equation __does not represent__ \( y \) as a function of \( x \).

c)

Solve the given equation for \( y \) to obtain: \( y = \dfrac{1}{x} \). The given equation __represents__ \( y \) as a function of \( x \) because for each value of \( x \), we have one value of \( y \) only.

d)

Given equation that \( x + y^2 = 25 \), solve for \( x \) to obtain \( x = 25 - y^2 = \).

Let \( y = 3 \), \( x = 25 - (3)^2 = 16 \).

Let \( y = - 3 \), \( x = 25 - ( - 3)^2 = 16 \).

Therefore for one input \( x = 16 \), we have two outputs \( y = - 3 \) and \( y = 3 \) and therefore the given equation __does not represent__ \( y \) as a function of \( x \).

The equation \( y = -2x + 6 \) represent \( y \) as a function of \( x \). A function can be given a name.

The function given by the above equation may be written as: \( y = f(x) \) where \( f \) is the name of the function and \( f(x) = -2x + 6 \).

\( x \) is the argument or input of the function, and \( y = f(x) \) is the value or output of the function. \( f(x) \) is also called the image of \( x \) by \( f \).

\( f(x) \) is read as " \( f \) of \( x \) ".

Hence, \( f( \color{red}2) = -2 \color{red}{(2)} + 6 = - 4 + 6 = 2 \)

Find the range of the given function below given its domain \( D \).

\( y = x^2 - 1 \) , \( D = \{-1 , 3 , 5 \} \)

Which of the following equations represents or does not represent \( y \) as a function of \( x \)? Explain why.

a) \( -3 x + y = x + 3 \)

b) \( | x | + 5 = y \)

c) \( \dfrac{y - 2}{x} = - 3 \)

d) \( x + | y | = 25 \)

Find the domain of the functions.

a) \(- y = x + 2 \)

b) \( 3x = y + 4 \)

c) \( y = \dfrac{1}{x - 2} \)

d) \( y = \dfrac{1}{\sqrt x} \)

Evaluate, if possible, the following.

\( f(0) \; , \; f(-2) \; , \; f(10) \; , \; g(-4) \) and \( \; g(-3) \) given that \( f(x) = -6 x + 10 \) and \( g(x) = \dfrac{1}{x+3} \).

a)

Substitute the value of \( x \) to find the corresponding value of \( y \).

For \( x = \color{red}{- 1} \) , \( \quad y = x^2 - 1 = \color{red}{(-1)}^2 - 1 = 1 - 1 = 0 \)

For \( x = \color{red}{3} \) , \( \quad y = x^2 - 1 = \color{red}{(3)}^2 - 1 = 9 - 1 = 8 \)

For \( x = \color{red}{5} \) , \( \quad y = x^2 - 1 = \color{red}{(5)}^2 - 1 = 25 - 1 = 24 \)

Range: \( \quad R = \{ 0 , 8 , 24 \} \)

a)

Solve the given equation for \( y \) and group like terms to obtain: \( \quad y = 4 x + 3 \)

For each value of the independent variable \( x \), there is one value only of the dependent variable \( y \) and hence the given equation represents \( y \) as a function of \( x \).

b)

The given equation may be written as: \( \quad y = | x | + 5 \)

For each value of the variable \( x \), there is one value only of the dependent variable \( y \) and hence the given equation represents \( y \) as a function of \( x \).

c)

Multiply both side of the given equation by \( x \) to obtain: \( \quad x \dfrac{y - 2}{x} = - 3 x\)

Simplify the above: \( \quad y - 2 = - 3x \)

Solve for \( y \): \( \quad y = - 3 x + 2 \)

For each value of the variable \( x \), there is one value only of the dependent variable \( y \) and hence the given equation represents \( y \) as a function of \( x \).

d)

Rewrite the given equation as: \( | y | = 25 - x \)

Substitute \( x \) by any real value in the above equation. For \( x = 4 \), for example, the above equation is written as. \( \quad | y | = 25 - 4 = 21\)

Solve the equation \( | y | = 21\) to obtain two solutions: \( y = 21 \) and \( y = -21 \)

Hence for the input \( x = 4 \), we have two values of the output \( y \) and therefore the given equation does not represent \( y \) as a function of \( x \).

a)

Solve the given equation for \( y \): \( \quad y = - x - 2 \)

The independent variable \( x \) in the above function may take any value if the set of the real numbers. Hence the __domain is the set of all real numbers__.

b)

Solve the given equation for \( y \): \( \quad y = 3 x - 4 \)

The independent variable \( x \) in the above function may take any value if the set of the real numbers. Hence the __domain is the set of all real numbers__.

c)

In the given function \( y = \dfrac{1}{x - 2} \), there is a division and a division by zero in not allowed in mathematics.
The denominator of the expression \( \dfrac{1}{x - 2} \) is \( x - 2 \) which is equal to zero for \( x = 2 \). Hence the __domain is the set of all real numbers except \( x = 2\)__.

d)

In the given function \( y = \dfrac{1}{\sqrt x} \), there is square root and a division. The square root is defined as a real number for \( x \) non negative. The division is not allowed for \( \sqrt x = 0 \) or \( x = 0 \).
Therefore the __domain of the given function is the set of all positive real numbers__.

Calculate by substitution.

\( f (\color{red}{0}) = -6 \color{red}{(0)} + 10 = 0 + 10 = 10 \)

\( f (\color{red}{-2}) = -6 \color{red}{(-2)} + 10 = 12 + 10 = 22 \)

\( f (\color{red}{10}) = -6 \color{red}{(10)} + 10 = -60 + 10 =-50 \)

\( g (\color{red}{-4}) = \dfrac{1}{(-4)+3} = \dfrac{1}{-1} = -1 \)

\( g (\color{red}{-3}) = \dfrac{1}{(-3)+3} = \dfrac{1}{0} = \text{undefined} \)

Algebra and Trigonometry - Swokowsky Cole - 1997 - ISBN: 0-534-95308-5

Algebra and Trigonometry with Analytic Geometry - R.E.Larson , R.P. Hostetler , B.H. Edwards, D.E. Heyd - 1997 - ISBN: 0-669-41723-8

Functions in Mathematics

domain and range .

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