FINDING DOMAIN AND RANGE OF A FUNCTION IN INTERVAL NOTATION

The domain of a function is the set of values that we are allowed to plug into our function.

The range of a function is the set of values that the function assumes. This set is the values that the function shoots out after we plug an x value in. They are the y values.

To represent the answer in interval notation, we should be aware of brackets.

( )  Open bracket

[  ]  Close bracket

Interval Notation

(-5, 3)

[-5, 3]

[-5, 3)

Possible Values

-4, -3, -2, -1, 0, 1, 2

-5, -4, -3, -2, -1, 0, 1, 2, 3

-5, -4, -3, -2, -1, 0, 1, 2 

Find the domain and range range of each of the following functions. Express answers in interval notation.

Problem 1 :

f(x)=3x+2

Solution:

f(x)=3x+2

For what value of x the function is defined, first we have to check for what values of x the function is undefined.

For that, we have to equate the denominator to 0. 

x + 2 = 0

x = -2

The domain is all real values except -2.

 x ≠ -2

Domain:

(-∞, -2) ∪ (-2, ∞)

Range:

g(x)=3x+2y=3x+2x+2=3yx=3y-2x=3-2yySubstitute f-1 for x and x for yf-1=3-2xxx0Range:(-,0)(0,)

Problem 2 :

g(x)=x+5x-2

Solution:

g(x)=x+5x-2

For what value of x the function is defined, first we have to check for what values of x the function is undefined.

For that, we have to equate the denominator to 0. 

x - 2 = 0

x = 2

The domain is all real values except 2.

 x ≠ 2

Domain:

(-∞, 2) ∪ (2, ∞)

Range:

g(x)=x+5x-2y=x+5x-2(x-2)y=x+5xy-2y=x+5xy-x=5+2yx(y-1)=5+2yx=5+2yy-1Substitute f-1 for x and x for yf-1=5+2xx-1x1Range: (-,1)(1,)

Problem 3 :

f(x)=4x-3

Solution:

f(x)=4x-3

For what value of x the function is defined, first we have to check for what values of x the function is undefined.

For that, we have to equate the denominator to 0. 

x - 3 = 0

x = 3

The domain is all real values except 3.

 x ≠ 3

Domain:

(-∞, 3) ∪ (3, ∞)

Range:

f(x)=4x-3y=4x-3x-3=4yx=4y+3x=4+3yySubstitute f-1 for x and x for y.f-1=4+3xxx0Range: (-,0)(0,)

Problem 4 :

g(x)=5x-2x-3

Solution:

g(x)=5x-2x-3

For what value of x the function is defined, first we have to check for what values of x the function is undefined.

For that, we have to equate the denominator to 0. 

x - 3 = 0

x = 3

The domain is all real values except 3.

 x ≠ 3

Domain:

(-∞, 3) ∪ (3, ∞)

Range:

g(x)=5x-2x-3y=5x-2x-3(x-3)y=5x-2xy-3y=5x-2xy-5x=3y-2x(y-5)=3y-2x=3y-2y-5Substitute f-1 for x and x for y.f-1=3x-2x-5x5Range:(-,5)(5,)

Problem 5 :

g(t) = |2x - 7| + 5

Solution:

g(t) = |2x - 7| + 5

Domain of g(t) is defined for all real values of x.

Domain:

(- ∞, ∞)

Range:

By analyzing the vertex of the absolute value function, we can figure out the range easily.

Vertex is at (7, 5). So, the range is [5, ∞)

Problem 6 :

h(t) = 6- |t - 1| 

Solution:

h(t) = 6 - |t - 1| 

h(t) = - |t - 1| + 6

The absolute value function opens down.

Domain of h(t) is defined for all real values of t.

Domain:

(- ∞, ∞)

Range:

h(t) = - |t - 1| + 6

Vertex is at (1, 6). So, the range is (-∞, 6].

Problem 7 :


f(x)=35x+6-4

Solution:

f(x)=35x+6-4

Domain:

The domain of function f defines by f(x) is the set of all real numbers. 

Domain: (-∞, ∞)

Range:

The range of function is the set of all real numbers.

Range: (-∞, ∞)

Problem 6 :

g(x)=48-3x+2

Solution:

g(x)=48-3x+2

Domain:

8-3x0-3x-8x83Domain:-,83

Range:

[2, ∞)

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