# FIND RANGE OF A FUNCTION FROM THE GIVEN DOMAIN

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.

Find the range for the functions with domain D :

Problem 1 :

D = {-1, 0, 2, 7, 9}, function: ‘add 3’.

Solution :

Let us create the function for the given rule.

f(x) = x + 3

D = {-1, 0, 2, 7, 9}

 If x = -1f(-1) = -1 + 3f(-1) = 2 if x = 0f(0) = 0 + 3f(0) = 3 if x = 2f(2) = 2 + 3f(2) = 5 if x = 7f(7) = 7 + 3f(7) = 10 if x = 9f(9) = 9 + 3f(9) = 12

Range = {2, 3, 5, 10, 12}

Problem 2 :

D = {-2, -1, 0, 1, 2}, function: ‘square and then divide by 2’.

Solution :

Let us create the function for the given rule.

f(x) = x²/2

D = {-2, -1, 0, 1, 2}

 If x = -2f(-2) = (-2)²/2= 4/2f(-2) = 2 if x = -1f(-1) = (-1)²/2f(-1) = 1/2 if x = 0f(0) = 0/2f(0) = 0 if x = 1f(1) = (1)²/2f(1) = 1/2 if x = 2f(2) = (2)²/2= 4/2f(2) = 2

Range = {2, 1/2, 0, 1/2, 2}

Problem 3 :

D = {x | -2 < x < 2}, function: ‘multiply x by 2 then add 1’.

Solution :

f(x) = 2x + 1

D = {-1, 0, 1}

 If x = -1f(-1) = 2(-1) + 1= -2 + 1f(-1)  = -1 if x = 0f(0) = 0 + 1f(0) = 1 if x = 1= 2 + 1f(1) = 3

Range = {-1, 1, 3}

Problem 4 :

D = {x | -3 ≤ x ≤ 4}, function: ‘cube x’.

Solution :

f(x) = x³

D = {-3, -2, -1, 0, 1, 2, 3, 4}

If x = -3 ==> f(-3) = (-3)3  ==>  -27

If x = -2 ==> f(-2) = (-2)3  ==>  -8

If x = -1 ==> f(-1) = (-1)3  ==>  -1

If x = 0 ==> f(0) = (0)3  ==>  0

If x = 1 ==> f(1) = (1)3  ==>  1

If x = 2 ==> f(2) = 23  ==>  8

If x = 3 ==> f(3) = 33  ==>  27

If x = 4 ==> f(4) = 43  ==>  64

Range = {-27, 8, -1, 0, 1, 8, 27, 64}

Problem 2 :

For the domain {0, 1, 2, 3} and the function ‘subtract 2’, find the range.

Solution :

f(x) = x - 2

D = {0, 1, 2, 3}

If x = 0 ==> f(0) = 0 - 2  ==>  -2

If x = 1 ==> f(1) = 1 - 2  ==>  -1

If x = 2 ==> f(2) = 2 - 2  ==>  0

If x = 3 ==> f(3) = 3 - 2  ==>  1

Range = {-2, -1, 0, 1}

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