Definition: Function A function is a rule for determining the value of one variable from the values of one or more other variables. We say that the first variable is a function of the other variable(s).
Figure 1.3 R1
R2
So D is a function of R and T. We also say that D= RT expresses D as a function of R and T. We will see rules expressed in other ways besides formulas when we study functions in greater depth in Chapter 2.
Formulas and functions are used to model real-life situations. The function D =RT models the relationship between distance, rate, and time in uniform motion.
The formula P = 2L + 2W models the relationship between the perimeter, length, and width of a rectangle. Since P is determined by the values of L and W, we say that P is a function of L and W. A list of common formulas that are commonly used as math- ematical models is given inside the back cover of this text.
The formula C =591F - 322 is used to determine the Celsius temperature from the Fahrenheit temperature. So C is a function of F. This formula is solved for C.
When a formula is solved for a specified variable, that variable is isolated on one side of the equal sign and must not occur on the other side. The formula F = 95 C + 32 expresses F as a function of C. It is solved for F.
EXAMPLE 1 Solving a formula for a specified variable
Solve the formula S= P + Prt for P.
Solution
P +Prt = S Write the formula with P on the left.
P11 + rt2 = S Factor out P.
P = S
1+ rt Divide each side by 1 +rt.
The formula S = P +Prt expresses S as a function of P, r, and t. The formula P = 1+Srt expresses P as a function of S, r, and t.
TRY THIS. Solve A = 12 hb1 + 12 hb2 for h. ■
In some situations we know the values of all variables except one. After we substitute values for those variables, the formula is an equation in one variable. We can then solve the equation to find the value of the remaining variable.
EXAMPLE 2 Finding the value of a variable in a formula
The total resistance R (in ohms) in the parallel circuit shown in Fig. 1.3 is modeled by the formula
1 R = 1
R1 + 1 R2.
The subscripts 1 and 2 indicate that R1 and R2 represent the resistance for two dif- ferent receivers. If R =7 ohms and R1 = 10 ohms, then what is the value of R2? Solution
Substitute the values for R and R1 and solve for R2. 1
7 = 1 10 + 1
R2 The LCD of 7, 10, and R2 is 70R2. 70R2#1
7 = 70R2 a1 10 + 1
R2
b Multiply each side by 70R2.
10R2 = 7R2+ 70
3R2 = 70 R2 = 70
3
Check the solution in the original formula. The resistance R2 is 70>3 ohms, or about 23.3 ohms.
TRY THIS. Find b1 if A =20, h =2, b2 = 3, and A = 12 h1b1 + b22. ■
Constructing Your Own Models
Applied problems in mathematics often involve solving an equation. The equations for some problems come from known formulas, as in Example 2, while in other problems we must write an equation that models a particular problem situation.
The best way to learn to solve problems is to study a few examples and then solve lots of problems. We will first look at an example of problem solving, then give a problem-solving strategy.
EXAMPLE 3 Solving a problem involving sales tax
Jeannie Fung bought a Ford Mustang GT for a total cost of $18,966, including sales tax. If the sales tax rate is 9%, then what amount of tax did she pay?
Solution
There are two unknown quantities here, the price of the car and the amount of the tax.
Represent the unknown quantities as follows:
x =the price of the car 0.09x =the amount of sales tax
The price of the car plus the amount of sales tax is the total cost of the car. We can model this relationship with an equation and solve it for x:
x + 0.09x = 18,966 1.09x = 18,966 x = 18,966
1.09 = 17,400 0.09x = 1566
You can check this answer by adding $17,400 and $1566 to get $18,966, the total cost. The amount of tax that Jeannie Fung paid was $1566.
TRY THIS. Joe bought a new computer for $1506.75, including sales tax at 5%.
What amount of tax did he pay? ■
No two problems are exactly alike, but there are similarities. The following strategy will assist you in solving problems on your own.
S T R A T E G Y Problem Solving
1. Read the problem as many times as necessary to get an understanding of the problem.
2. If possible, draw a diagram to illustrate the problem.
3. Choose a variable, write down what it represents, and if possible, rep- resent any other unknown quantities in terms of that variable.
In the next example we have a geometric situation. Notice how we are follow- ing the strategy for problem solving.
EXAMPLE 4 Solving a geometric problem
In 1974, Chinese workers found three pits that contained life-size sculptures of war- riors, which were created to guard the tomb of an emperor (www.chinatour.com).
The largest rectangular pit has a length that is 40 yards longer than three times the width. If its perimeter is 640 yards, what are the length and width?
Solution
First draw a diagram as shown in Fig. 1.4. Use the fact that the length is 40 yards longer than three times the width to represent the width and length as follows:
x= the width in yards 3x+ 40 = the length in yards
4. Write an equation that models the situation. You may be able to use a known formula, or you may have to write an equation that models only that particular problem.
5. Solve the equation.
6. Check your answer by using it to solve the original problem (not just the equation).
7. Answer the question posed in the original problem.
The formula for the perimeter of a rectangle is 2L +2W = P. Replace W by x, L by 3x + 40, and P by 640.
2L + 2W = P Perimeter formula 213x + 402 +2x = 640 Substitution
6x +80 +2x = 640 8x + 80 = 640 8x = 560 x = 70
If x = 70, then 3x + 40 = 250. Check that the dimensions of 70 yards and 250 yards give a perimeter of 640 yards. We conclude that the length of the pit is 250 yards and its width is 70 yards.
TRY THIS. The length of a rectangle is 20 cm shorter than five times its width. If
the perimeter is 800 cm, then what is the length? ■
3x + 40
x
Figure 1.4
The next problem is called a uniform-motion problem because it involves motion at a constant rate. Of course, people do not usually move at a constant rate, but their average speed can be assumed to be constant over some time interval. The problem also illustrates how a table can be used as an effective technique for orga- nizing information.
EXAMPLE 5 Solving a uniform-motion problem
A group of hikers from Tulsa hiked down into the Grand Canyon in 3 hours 30 minutes.
Coming back up on a trail that was 4 miles shorter, they hiked 2 mph slower and it took them 1 hour longer. What was their rate going down?
Solution
Let x represent the rate going down into the canyon. Make a table to show the dis- tance, rate, and time for both the trip down and the trip back up. Once we fill in any two entries in a row of the table, we can use D = RT to obtain an expression for the third entry:
Rate Time Distance
Down x mi>hr 3.5 hr 3.5x mi
Up x -2 mi>hr 4.5 hr 4.51x- 22 mi
Using the fact that the distance up was 4 miles shorter, we can write the following equation:
4.51x -22 = 3.5x - 4 4.5x -9 = 3.5x - 4
x -9 = -4 x = 5
After checking that 5 mph satisfies the conditions given in the problem, we con- clude that the hikers traveled at 5 mph going down into the canyon.
TRY THIS. Bea hiked uphill from her car to a waterfall in 6 hours. Hiking back to her car over the same route, she averaged 2 mph more and made the return trip in
half the time. How far did she hike? ■
Average speed is not necessarily the average of your speeds. For example, if you drive 70 mph for 3 hours and then 30 mph for 1 hour, you will travel 240 miles in 4 hours. Your average speed for the trip is 60 mph, which is not the average of 70 and 30.
EXAMPLE 6 Average speed
Shelly drove 40 miles from Peoria to Bloomington at 20 mph. She then drove back to Peoria at a higher rate of speed so that she averaged 30 mph for the whole trip.
What was her speed on the return trip?
Solution
Let x represent the speed on the return trip. Make a table as follows:
Rate Time Distance
Going 20 mph 2 hr 40 mi
Returning x mph 40>x hr 40 mi Round trip 30 mph 2 +40>x hr 80 mi
Use D= RT to write an equation for the round trip and solve it.
80 = 30a2 + 40 xb 80 = 60 + 1200
x 20 = 1200
x 20x = 1200
x = 60
So the speed on the return trip was 60 mph. It is interesting to note that the distance between the cities does not affect the solution. You should repeat this example using d as the distance between the cities.
TRY THIS. Dee drove 20 miles to work at 60 mph. She then drove back home at a lower rate of speed, averaging 50 mph for the round trip. What was her speed on the
return trip? ■
The next example involves mixing beverages with two different concentrations of orange juice. In other mixture problems, we may mix chemical solutions, candy, or even people.
EXAMPLE 7 Solving a mixture problem
A beverage producer makes two products, Orange Drink, containing 10% orange juice, and Orange Delight, containing 50% orange juice. How many gallons of Orange Delight must be mixed with 300 gallons of Orange Drink to create a new product containing 40% orange juice?
Solution
Let x represent the number of gallons of Orange Delight and make a sketch as in Fig. 1.5. Next, we make a table that shows three pertinent expressions for each prod- uct: the amount of the product, the percent of orange juice in the product, and the actual amount of orange juice in that product.
Figure 1.5
300 gallons x gallons x ⫹ 300 gallons
⫹ ⫽
Amount of
Product
Percent Orange Juice
Amount Orange Juice Drink 300 gal 10% 0.1013002 gal
Delight x gal 50% 0.50x gal
Mixture x +300 gal 40% 0.401x +3002 gal We can now write an equation expressing the fact that the actual amount of orange juice in the mixture is the sum of the amounts of orange juice in the Orange Drink and in the Orange Delight:
0.401x +3002 = 0.1013002+ 0.50x 0.4x +120 = 30 + 0.5x
90 = 0.1x 900 = x
Mix 900 gallons of Orange Delight with the 300 gallons of Orange Drink to obtain the proper mixture.
TRY THIS. How many gallons of a 40% acid solution must be mixed with 30 gal- lons of a 20% acid solution to obtain a mixture that is 35% acid? ■
Work problems are problems in which people or machines are working to- gether to accomplish a task. A typical situation might have two people painting a house at different rates. Suppose Joe and Frank are painting a house together for 2 hours and Joe paints at the rate of one-sixth of the house per hour while Frank paints at the rate of one-third of the house per hour. Note that
a1
6 of house per hourb12 hr2 = 1 3 of house and
a1
3 of house per hourb12 hr2= 2
3 of house.
The product of the rate and the time gives the fraction of the house completed by each person, and these fractions have a sum of 1 because the entire job is completed.
Note how similar this situation is to a uniform-motion problem where RT = D.
EXAMPLE 8 Solving a work problem
Aboard the starship Nostromo, the human technician, Brett, can process the crew’s medical history in 36 minutes. However, the android Science Officer, Ash, can process the same records in 24 minutes. After Brett worked on the records for 1 minute, Ash joined in and both crew members worked until the job was done. How long did Ash work on the records?
Solution
Let x represent the number of minutes that Ash worked and x + 1 represent the number of minutes that Brett worked. Ash works at the rate of 1>24 of the job per minute, while Brett works at the rate of 1>36 of the job per minute. The following table shows all of the pertinent quantities.
Rate Time Work Completed
Ash 1
24 job>min x min 1
24x job Brett 1
36 job>min x+ 1 min 1
361x +12 job
The following equation expresses the fact that the work completed together is the sum of the work completed by each worker alone.
1 24 x+ 1
36 1x + 12 = 1 72J1
24 x + 1
36 1x+ 12j = 72#1 Multiply by the LCD 72.
3x +2x + 2= 72 5x = 70 x = 14 Ash worked for 14 minutes.
Check 14 in the original equation as shown in Fig. 1.6.
TRY THIS. A tank can be filled by a small pipe in 12 hours or a large pipe in 8 hours.
How long will it take to fill the tank if the small pipe is used alone for 2 hours and
then both pipes are used until the tank is full? ■
Figure 1.6
1. If we solve P + Prt= S for P, we get P = S- Prt.
2. The perimeter of any rectangle is the product of its length and width.
3. If n is an odd integer, then n +1 and n +3 represent odd integers.
4. The equation y= x + 2 expresses y as a function of x.
5. Two numbers that have a sum of -3 can be represented by x and -3 -x.
6. If P is the number of professors and S is the number of students at the play, and there are twice as many professors as students, then 2P = S.
7. If you need $100,000 for your house, and the agent gets 9% of the selling price, then the agent gets $9000, and the house sells for $109,000.
8. If John can mow the lawn in x hours, then he mows the lawn at the rate of 1>x of the lawn per hour.
9. If George hiked 3x miles and Anita hiked 41x -22 miles, and George hiked 5 more miles than Anita, then 3x + 5= 41x- 22.
10. Two numbers that differ by 9 can be represented as 9 and x+ 9.
FOR thought... True or False? Explain.
Fill in the blank.
1. A ________ is an equation that involves two or more variables.
2. If the value of y can be determined from the value of x, then y is a ________ of x.
3. Motion at a constant rate is ________ motion.
4. In uniform motion, distance is a function of ________ and ________ .
Solve each formula for the specified variable. The use of the formula is indicated in parentheses.
5. I= Prt for r (simple interest) 6. D=RT for R (uniform motion) 7. F= 9
5 C+32 for C (temperature) 8. C= 5
9 1F -322 for F (temperature) 9. A = 1
2 bh for b (area of a triangle) 10. A = 1
2 bh for h (area of a triangle) 11. Ax+By= C for y (equation of a line) 12. Ax+By= C for x (equation of a line) 13. 1
R = 1 R1 + 1
R2 + 1
R3 for R1 (resistance)
14. 1 R = 1
R1 + 1 R2 + 1
R3 for R2 (resistance) 15. an= a1 +1n -12d for n (arithmetic sequence) 16. Sn= n
2 1a1+ an2 for a1 (arithmetic series) 17. S= a1- a1 rn
1 -r for a1 (geometric series) 18. S= 2LW+ 2LH+ 2HW for H (surface area) 19. The 2.4-Meter Rule A 2.4-meter sailboat is a one-person
boat that is about 13 ft in length, has a displacement of about 550 lb, and a sail area of about 81 ft2. To compete in the 2.4-meter class, a boat must satisfy the formula