Download Function Notation: Understanding Inputs and Outputs of Functions and more Summaries Algebra in PDF only on Docsity!
44 Relations and Functions
1.5 Function Notation
In Definition 1.4, we described a function as a special kind of relation − one in which each x- coordinate is matched with only one y-coordinate. In this section, we focus more on the process by which the x is matched with the y. If we think of the domain of a function as a set of inputs and the range as a set of outputs, we can think of a function f as a process by which each input x is matched with only one output y. Since the output is completely determined by the input x and the process f , we symbolize the output with function notation: ‘f (x)’, read ‘f of x.’ In this case, the parentheses here do not indicate multiplication, as they do elsewhere in algebra. This could cause confusion if the context is not clear. In other words, f (x) is the output which results by applying the process f to the input x. This relationship is typically visualized using a diagram similar to the one below.
f
x Domain (Inputs)
y = f (x) Range (Outputs)
The value of y is completely dependent on the choice of x. For this reason, x is often called the independent variable, or argument of f , whereas y is often called the dependent variable.
As we shall see, the process of a function f is usually described using an algebraic formula. For example, suppose a function f takes a real number and performs the following two steps, in sequence
- multiply by 3
- add 4
If we choose 5 as our input, in step 1 we multiply by 3 to get (5)(3) = 15. In step 2, we add 4 to our result from step 1 which yields 15 + 4 = 19. Using function notation, we would write f (5) = 19 to indicate that the result of applying the process f to the input 5 gives the output 19. In general, if we use x for the input, applying step 1 produces 3x. Following with step 2 produces 3x + 4 as our final output. Hence for an input x, we get the output f (x) = 3x + 4. Notice that to check our formula for the case x = 5, we replace the occurrence of x in the formula for f (x) with 5 to get f (5) = 3(5) + 4 = 15 + 4 = 19, as required.
1.5 Function Notation 45
Example 1.5.1. Suppose a function g is described by applying the following steps, in sequence
- add 4
- multiply by 3
Determine g(5) and find an expression for g(x).
Solution. Starting with 5, step 1 gives 5 + 4 = 9. Continuing with step 2, we get (3)(9) = 27. To find a formula for g(x), we start with our input x. Step 1 produces x + 4. We now wish to multiply this entire quantity by 3, so we use a parentheses: 3(x + 4) = 3x + 12. Hence, g(x) = 3x + 12. We can check our formula by replacing x with 5 to get g(5) = 3(5) + 12 = 15 + 12 = 27 X.
Most of the functions we will encounter in College Algebra will be described using formulas like the ones we developed for f (x) and g(x) above. Evaluating formulas using this function notation is a key skill for success in this and many other math courses.
Example 1.5.2. For f (x) = −x^2 + 3x + 4, find and simplify
- f (−1), f (0), f (2)
- f (2x), 2f (x)
- f (x + 2), f (x) + 2, f (x) + f (2)
Solution.
- To find f (−1), we replace every occurrence of x in the expression f (x) with − 1
f (−1) = −(−1)^2 + 3(−1) + 4 = −(1) + (−3) + 4 = 0
Similarly, f (0) = −(0)^2 + 3(0) + 4 = 4, and f (2) = −(2)^2 + 3(2) + 4 = −4 + 6 + 4 = 6.
- To find f (2x), we replace every occurrence of x with the quantity 2x
f (2x) = −(2x)^2 + 3(2x) + 4 = −(4x^2 ) + (6x) + 4 = − 4 x^2 + 6x + 4
The expression 2f (x) means we multiply the expression f (x) by 2
2 f (x) = 2
−x^2 + 3x + 4
= − 2 x^2 + 6x + 8
1.5 Function Notation 47
As long as we substitute numbers other than 3 and −3, the expression r(x) is a real number. Hence, we write our domain in interval notation as (−∞, −3) ∪ (− 3 , 3) ∪ (3, ∞). When a formula for a function is given, we assume the function is valid for all real numbers which make arithmetic sense when substituted into the formula. This set of numbers is often called the implied domain^1 of the function. At this stage, there are only two mathematical sins we need to avoid: division by 0 and extracting even roots of negative numbers. The following example illustrates these concepts.
Example 1.5.3. Find the domain^2 of the following functions.
- f (x) =
4 x x − 3
- g(x) =
4 − 3 x
- h(x) = 5
4 − 3 x
- r(x) =
x + 3
- I(x) =
3 x^2 x
Solution.
- In the expression for f , there are two denominators. We need to make sure neither of them is
- To that end, we set each denominator equal to 0 and solve. For the ‘small’ denominator, we get x − 3 = 0 or x = 3. For the ‘large’ denominator
4 x x − 3
4 x x − 3 (1)(x − 3) =
4 x x − 3
(x^ −^ 3)^ clear denominators
x − 3 = 4 x
− 3 = 3 x
− 1 = x
So we get two real numbers which make denominators 0, namely x = −1 and x = 3. Our domain is all real numbers except −1 and 3: (−∞, −1) ∪ (− 1 , 3) ∪ (3, ∞).
(^1) or, ‘implicit domain’ (^2) The word ‘implied’ is, well, implied.
48 Relations and Functions
- The potential disaster for g is if the radicand^3 is negative. To avoid this, we set 4 − 3 x ≥ 0
4 − 3 x ≥ 0 4 ≥ 3 x 4 3 ≥ x
Hence, as long as x ≤ 43 , the expression 4 − 3 x ≥ 0, and the formula g(x) returns a real number. Our domain is
]
- The formula for h(x) is hauntingly close to that of g(x) with one key difference − whereas the expression for g(x) includes an even indexed root (namely a square root), the formula for h(x) involves an odd indexed root (the fifth root.) Since odd roots of real numbers (even negative real numbers) are real numbers, there is no restriction on the inputs to h. Hence, the domain is (−∞, ∞).
- To find the domain of r, we notice that we have two potentially hazardous issues: not only do we have a denominator, we have a square root in that denominator. To satisfy the square root, we set the radicand x + 3 ≥ 0 so x ≥ −3. Setting the denominator equal to zero gives
x + 3 = 0 6 =
x + 3 62 =
x + 3
36 = x + 3 33 = x
Since we squared both sides in the course of solving this equation, we need to check our answer. Sure enough, when x = 33, 6 −
x + 3 = 6 −
36 = 0, and so x = 33 will cause problems in the denominator. At last we can find the domain of r: we need x ≥ −3, but x 6 = 33. Our final answer is [− 3 , 33) ∪ (33, ∞).
- It’s tempting to simplify I(x) = 3 x 2 x = 3x, and, since there are no longer any denominators, claim that there are no longer any restrictions. However, in simplifying I(x), we are assuming x 6 = 0, since 00 is undefined.^4 Proceeding as before, we find the domain of I to be all real numbers except 0: (−∞, 0) ∪ (0, ∞).
It is worth reiterating the importance of finding the domain of a function before simplifying, as evidenced by the function I in the previous example. Even though the formula I(x) simplifies to 3 x, it would be inaccurate to write I(x) = 3x without adding the stipulation that x 6 = 0. It would be analogous to not reporting taxable income or some other sin of omission.
(^3) The ‘radicand’ is the expression ‘inside’ the radical. (^4) More precisely, the fraction 0 0 is an ‘indeterminant form’.^ Much time will be spent in Calculus wrestling with such creatures.
