# Quantitative Aptitude > Functions and applications

### Functions

• An algebraic expression in one variable can be used to define a function of that variable. Functions are usually denoted by letters such as f, g, and h. For example, the algebraic expression 3x_5 can be used to define a function f by

where f (x) is called the value of f at x and is obtained by substituting the value of x in the expression above. For example, if x_1 is substituted in the expression above, the result is .

• It might be helpful to think of a function f as a machine that takes an input, which is a value of the variable x, and produces the corresponding output, f (x). For any function, each input x gives exactly one output f (x). However, more than one value of x can give the same output f (x). For example, if g is the function defined by , then  and .
• The domain of a function is the set of all permissible inputs, that is, all permissible values of the variable x. For the functions f and g defined above, the domain is the set of all real numbers.

Ex. Let f be the function defined by . In this case, f is not defined at   because  is not defined. Hence, the domain of f consists of all real numbers except for 6.

Ex. Let h be the function defined by , the absolute value of , which is the distance between  and 0 on the number line. The domain of  is the set of all real numbers. Also,  for all real numbers , which reflects the property that on the number line the distance between  and 0 is the same as the distance between  and 0.

### Applications

• Translating verbal descriptions into algebraic expressions is an essential initial step in solving word problems. Some examples are given below.

Ex. Ellen has received the following scores on 3 exams: 82, 74, and 90. What score will Ellen need to receive on the next exam so that the average (arithmetic mean) score for the 4 exams will be 85?

Sol. Let x represent the score on Ellen's next exam. This initial step of assigning a variable to the quantity that is sought is an important beginning to solving the problem. Then in terms of , the average of the 4 exams is

Solving the resulting linear equation for x, you get

Therefore, Ellen will need to attain a score of 94 on the next exam.

Ex.  A mixture of 12 ounces of vinegar and oil is 40 percent vinegar, where all of the measurements are by weight. How many ounces of oil must be added to the mixture to produce a new mixture that is only 25 percent vinegar?

Sol. Let  represent the number of ounces of oil to be added. Then the total number of ounces of the new mixture will be , and the total number of ounces of vinegar in the new mixture will be . Since the new mixture must be 25 percent vinegar,

Therefore,

Thus, 7.2 ounces of oil must be added to produce a new mixture that is 25 percent vinegar.

Ex. To produce a particular radio model, it costs a manufacturer \$30 per radio, and it is assumed that if 500 radios are produced, all of them will be sold. What must be the selling price per radio to ensure that the profit (revenue from the sales minus the total production cost) on the 500 radios is greater than \$8,200?

Sol. If y represents the selling price per radio, then the profit is 500(y – 30). Therefore, we set

Solving the inequality, we get

Thus, the selling price must be greater than \$46.40 to ensure that the profit is greater than \$8,200.

• Some applications involve computing interest earned on an investment during a specified time period. The interest can be computed as simple interest or compound interest.
• Simple interest is based only on the initial deposit, which serves as the amount on which interest is computed, called the principal, for the entire time period. If the amount P is invested at a simple annual interest rate of r percent, then the value V of the investment at the end of t years is given by the formula

where P and V are in dollars.

• In the case of compound interest, interest is added to the principal at regular time intervals, such as annually, quarterly, and monthly. Each time interest is added to the principal, the interest is said to be compounded. After each compounding, interest is earned on the new principal, which is the sum of the preceding principal and the interest just added. If the amount P is invested at an annual interest rate of r percent, compounded annually, then the value V of the investment at the end of t years is given by the formula

If the amount P is invested at an annual interest rate of r percent, compounded n times per year, then the value V of the investment at the end of t years is given by the formula

Ex. If \$10,000 is invested at a simple annual interest rate of 6 percent, what is the value of the investment after half a year?

Sol. According to the formula for simple interest, the value of the investment after 1 year is

Ex. If an amount P is to be invested at an annual interest rate of 3.5 percent, compounded annually, what should be the value of P so that the value of the investment is \$1,000 at the end of 3 years?

Sol. According to the formula for 3.5 percent annual interest, compounded annually, the value of the investment after 3 years is

And we set it equal to \$1,000

To find the value of P, we divide both sides of the equation by

Thus, an amount of approximately \$901.94 should be invested.

Ex. A college student expects to earn at least \$1,000 in interest on an initial investment of \$20,000. If the money is invested for one year at interest compounded quarterly, what is the least annual interest rate that would achieve the goal?

Sol. According to the formula for r percent annual interest, compounded quarterly, the value of the investment after 1 year is

By setting this value greater than or equal to \$21,000 and solving for r, we get

We can use the fact that taking the positive fourth root of each side of an inequality preserves the direction of the inequality. This is also true for the positive square root or any other positive root.

To compute the fourth root, we can use the fact that  for ; that is, we can compute a fourth root by taking a square root twice:

So the least annual interest rate is approximately 4.91 percent.