FIND RANGE OF A FUNCTION FROM THE GIVEN THE DOMAIN

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}

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}

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}

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 5 :

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}

Problem 6 :

The domain of the function represented by 2x + y = 8 is −2, 0, 2, 4, and 6. What is the range of the function represented by the table?

Solution :

2x + y = 8

y = -2x + 8

So, the range is {12, 8, 4, 0, -4}.

Problem 7 :

The table shows the percent y (in decimal form) of the moon that was visible at midnight x days after January 24, 2011.

(a) Interpret the domain and range.

(b) What percent of the moon was visible on January 26, 2011?

Solution :

a)  Zero days after January 24 is January 24. One day after January 24 is January 25. So, the domain of 0, 1, 2, 3, and 4 represents January 24, 25, 26, 27 and 28.

The range is 0.76, 0.65, 0.54, 0.43, and 0.32. These amounts are decreasing, so the moon was less visible each day.

b. January 26, 2011 corresponds to the input x = 2. When x = 2, y = 0.54. So, 0.54, or 54% of the moon was visible on January 26, 2011

Problem 8 :

In the sport of vaulting, a vaulter performs a routine while on a moving horse. For each round x of competition, the vaulter receives a score y from 1 to 10.

a. Find the domain and range of the function represented by the table.

b. Interpret the domain and range.

c. What is the mean score of the vaulter?

Solution :

a) Domain = {1, 2, 3}

Range = {6.856, 7.923, 8.135}

b) Here x represents the number of rounds and y represents scores.

c) Mean

= (6.856 + 7.923 + 8.135)/3

= 7.638