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1.4 Function Notation 55
1.4 Function Notation
In Definition 1.6, 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 other words, f (x) is the output which results by applying the process f to the input x. In this case, the parentheses here do not indicate multiplication, as they do elsewhere in Algebra. This can cause confusion if the context is not clear, so you must read carefully. 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.
56 Relations and Functions
Example 1.4.1. 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 1.4.2. Let f (x) = −x^2 + 3x + 4
- Find and simplify the following.
(a) f (−1), f (0), f (2) (b) f (2x), 2f (x) (c) f (x + 2), f (x) + 2, f (x) + f (2)
- Solve f (x) = 4.
Solution.
- (a) 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. (b) 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
58 Relations and Functions
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^1 as (−∞, −3) ∪ (− 3 , 3) ∪ (3, ∞). When a formula for a function is given, we assume that 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^2 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 1.4.3. Find the domain^3 of the following functions.
- g(x) =
4 − 3 x 2. h(x) = 5
4 − 3 x
- f (x) =
4 x x − 3
- F (x) =
√ (^42) x + 1
x^2 − 1
- r(t) =
t + 3 6.^ I(x) =
3 x^2 x
Solution.
- The potential disaster for g is if the radicand^4 is negative. To avoid this, we set 4 − 3 x ≥ 0. From this, we get 3x ≤ 4 or x ≤ 43. What this shows is that 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 (−∞, ∞).
- 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
(^1) See the Exercises for Section 1.1. (^2) or, ‘implicit domain’ (^3) The word ‘implied’ is, well, implied. (^4) The ‘radicand’ is the expression ‘inside’ the radical.
1.4 Function Notation 59
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, ∞).
- In finding the domain of F , we notice that we have two potentially hazardous issues: not only do we have a denominator, we have a fourth (even-indexed) root. Our strategy is to determine the restrictions imposed by each part and select the real numbers which satisfy both conditions. To satisfy the fourth root, we require 2x + 1 ≥ 0. From this we get 2x ≥ − 1 or x ≥ − 12. Next, we round up the values of x which could cause trouble in the denominator by setting the denominator equal to 0. We get x^2 − 1 = 0, or x = ±1. Hence, in order for a real number x to be in the domain of F , x ≥ − 12 but x 6 = ±1. In interval notation, this set is
[
- Don’t be put off by the ‘t’ here. It is an independent variable representing a real number, just like x does, and is subject to the same restrictions. As in the previous problem, we have double danger here: we have a square root and a denominator. To satisfy the square root, we need a non-negative radicand so we set t + 3 ≥ 0 to get t ≥ −3. Setting the denominator equal to zero gives 6 −
t + 3 = 0, or
t + 3 = 6. Squaring both sides gives t + 3 = 36, or t = 33. Since we squared both sides in the course of solving this equation, we need to check our answer.^5 Sure enough, when t = 33, 6 −
t + 3 = 6 −
36 = 0, so t = 33 will cause problems in the denominator. At last we can find the domain of r: we need t ≥ −3, but t 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.^6 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
(^5) Do you remember why? Consider squaring both sides to ‘solve’ √t + 1 = −2. (^6) More precisely, the fraction 0 0 is an ‘indeterminant form’. Calculus is required tame such beasts.
1.4 Function Notation 61
these limits determine the applied domain of g. Typically, an applied domain is stated explicitly. In this case, it would be common to see something like c(g) = 1. 5 g, 0 ≤ g ≤ 100, meaning the number of pounds of grapes purchased is limited from 0 up to 100. The upper bound here, 100 may represent the inventory of the market, or some other limit as set by local policy or law. Even with this restriction, our model has its limitations. As we saw above, it is virtually impossible to buy exactly 3.3 pounds of grapes so that our cost is exactly $5. In this case, being sensible shoppers, we would most likely ‘round down’ and purchase 3 pounds of grapes or however close the market scale can read to 3.3 without being over. It is time for a more sophisticated example.
Example 1.4.4. The height h in feet of a model rocket above the ground t seconds after lift-off is given by
h(t) =
− 5 t^2 + 100t, if 0 ≤ t ≤ 20 0 , if t > 20
- Find and interpret h(10) and h(60).
- Solve h(t) = 375 and interpret your answers.
Solution.
- We first note that the independent variable here is t, chosen because it represents time. Secondly, the function is broken up into two rules: one formula for values of t between 0 and 20 inclusive, and another for values of t greater than 20. Since t = 10 satisfies the inequality 0 ≤ t ≤ 20, we use the first formula listed, h(t) = − 5 t^2 + 100t, to find h(10). We get h(10) = −5(10)^2 + 100(10) = 500. Since t represents the number of seconds since lift-off and h(t) is the height above the ground in feet, the equation h(10) = 500 means that 10 seconds after lift-off, the model rocket is 500 feet above the ground. To find h(60), we note that t = 60 satisfies t > 20, so we use the rule h(t) = 0. This function returns a value of 0 regardless of what value is substituted in for t, so h(60) = 0. This means that 60 seconds after lift-off, the rocket is 0 feet above the ground; in other words, a minute after lift-off, the rocket has already returned to Earth.
- Since the function h is defined in pieces, we need to solve h(t) = 375 in pieces. For 0 ≤ t ≤ 20, h(t) = − 5 t^2 + 100t, so for these values of t, we solve − 5 t^2 + 100t = 375. Rearranging terms, we get 5t^2 − 100 t + 375 = 0, and factoring gives 5(t − 5)(t − 15) = 0. Our answers are t = 5 and t = 15, and since both of these values of t lie between 0 and 20, we keep both solutions. For t > 20, h(t) = 0, and in this case, there are no solutions to 0 = 375. In terms of the model rocket, solving h(t) = 375 corresponds to finding when, if ever, the rocket reaches 375 feet above the ground. Our two answers, t = 5 and t = 15 correspond to the rocket reaching this altitude twice – once 5 seconds after launch, and again 15 seconds after launch.^12 (^12) What goes up...
62 Relations and Functions
The type of function in the previous example is called a piecewise-defined function, or ‘piecewise’ function for short. Many real-world phenomena (e.g. postal rates,^13 income tax formulas^14 ) are modeled by such functions.
By the way, if we wanted to avoid using a piecewise function in Example 1.4.4, we could have used h(t) = − 5 t^2 + 100t on the explicit domain 0 ≤ t ≤ 20 because after 20 seconds, the rocket is on the ground and stops moving. In many cases, though, piecewise functions are your only choice, so it’s best to understand them well.
Mathematical modeling is not a one-section topic. It’s not even a one-course topic as is evidenced by undergraduate and graduate courses in mathematical modeling being offered at many universities. Thus our goal in this section cannot possibly be to tell you the whole story. What we can do is get you started. As we study new classes of functions, we will see what phenomena they can be used to model. In that respect, mathematical modeling cannot be a topic in a book, but rather, must be a theme of the book. For now, we have you explore some very basic models in the Exercises because you need to crawl to walk to run. As we learn more about functions, we’ll help you build your own models and get you on your way to applying Mathematics to your world.
(^13) See the United States Postal Service website http://www.usps.com/prices/first-class-mail-prices.htm (^14) See the Internal Revenue Service’s website http://www.irs.gov/pub/irs-pdf/i1040tt.pdf
64 Relations and Functions
- f (x) = 2x + 1 12. f (x) = 3 − 4 x
- f (x) = 2 − x^2 14. f (x) = x^2 − 3 x + 2
- f (x) =
x x − 1 16.^ f^ (x) =^
x^3
- f (x) = 6 18. f (x) = 0
In Exercises 19 - 26, use the given function f to find and simplify the following:
- f (2) • f (−2) • f (2a)
- 2 f (a) • f (a + 2) • f (a) + f (2)
- f
a
- f^ ( 2 a ) •^ f^ (a^ +^ h)
- f (x) = 2x − 5 20. f (x) = 5 − 2 x
- f (x) = 2x^2 − 1 22. f (x) = 3x^2 + 3x − 2
- f (x) =
2 x + 1 24. f (x) = 117
- f (x) = x 2 26.^ f^ (x) =
x
In Exercises 27 - 34, use the given function f to find f (0) and solve f (x) = 0
- f (x) = 2x − 1 28. f (x) = 3 − 25 x
- f (x) = 2x^2 − 6 30. f (x) = x^2 − x − 12
- f (x) =
x + 4 32. f (x) =
1 − 2 x
- f (x) =
4 − x 34.^ f^ (x) =
3 x^2 − 12 x 4 − x^2
- Let f (x) =
√^ x^ + 5,^ x^ ≤ −^3 9 − x^2 , − 3 < x ≤ 3 −x + 5, x > 3
Compute the following function values.
(a) f (−4) (b) f (−3) (c) f (3)
(d) f (3.001) (e) f (− 3 .001) (f) f (2)
1.4 Function Notation 65
- Let f (x) =
x^2 if x ≤ − 1 √ 1 − x^2 if − 1 < x ≤ 1 x if x > 1
Compute the following function values.
(a) f (4) (b) f (−3) (c) f (1)
(d) f (0) (e) f (−1) (f) f (− 0 .999)
In Exercises 37 - 62, find the (implied) domain of the function.
- f (x) = x^4 − 13 x^3 + 56x^2 − 19 38. f (x) = x^2 + 4
- f (x) =
x − 2 x + 1
- f (x) =
3 x x^2 + x − 2
- f (x) =
2 x x^2 + 3
- f (x) =
2 x x^2 − 3
- f (x) = x + 4 x^2 − 36
- f (x) = x − 2 x − 2
- f (x) =
3 − x 46. f (x) =
2 x + 5
- f (x) = 9x
x + 3 (^) 48. f (x) =
7 − x x^2 + 1
- f (x) =
6 x − (^2) 50. f (x) = √^6 6 x − 2
- f (x) = 3
6 x − (^2) 52. f (x) = 6 4 −
6 x − 2
- f (x) =
6 x − 2 x^2 − 36
- f (x) =
√ (^36) x − 2
x^2 + 36
- s(t) = t t − 8 56.^ Q(r) =
r r − 8
- b(θ) = θ √ θ − 8
- A(x) =
x − 7 +
9 − x
- α(y) = 3
y y − 8
- g(v) =
v^2
- T (t) =
t − 8 5 − t
- u(w) = w − 8 5 −
w
1.4 Function Notation 67
(b) How much does it cost to order 50 copies of the book? What about 51 copies? (c) Your answer to 72b should get you thinking. Suppose a bookstore estimates it will sell 50 copies of the book. How many books can, in fact, be ordered for the same price as those 50 copies? (Round your answer to a whole number of books.)
- An on-line comic book retailer charges shipping costs according to the following formula
S(n) =
- 5 n + 2. 5 if 1 ≤ n ≤ 14 0 if n ≥ 15
where n is the number of comic books purchased and S(n) is the shipping cost in dollars.
(a) What is the cost to ship 10 comic books? (b) What is the significance of the formula S(n) = 0 for n ≥ 15?
- The cost C (in dollars) to talk m minutes a month on a mobile phone plan is modeled by
C(m) =
25 if 0 ≤ m ≤ 1000 25 + 0.1(m − 1000) if m > 1000
(a) How much does it cost to talk 750 minutes per month with this plan? (b) How much does it cost to talk 20 hours a month with this plan? (c) Explain the terms of the plan verbally.
- In Section 1.1.1 we defined the set of integers as Z = {... , − 3 , − 2 , − 1 , 0 , 1 , 2 , 3 ,.. .}.^15 The greatest integer of x, denoted by bxc, is defined to be the largest integer k with k ≤ x.
(a) Find b 0. 785 c, b 117 c, b− 2. 001 c, and bπ + 6c (b) Discuss with your classmates how bxc may be described as a piecewise defined function. HINT: There are infinitely many pieces! (c) Is ba + bc = bac + bbc always true? What if a or b is an integer? Test some values, make a conjecture, and explain your result.
- We have through our examples tried to convince you that, in general, f (a + b) 6 = f (a) + f (b). It has been our experience that students refuse to believe us so we’ll try again with a different approach. With the help of your classmates, find a function f for which the following properties are always true.
(a) f (0) = f (−1 + 1) = f (−1) + f (1) (^15) The use of the letter Z for the integers is ostensibly because the German word zahlen means ‘to count.’
68 Relations and Functions
(b) f (5) = f (2 + 3) = f (2) + f (3) (c) f (−6) = f (0 − 6) = f (0) − f (6) (d) f (a + b) = f (a) + f (b) regardless of what two numbers we give you for a and b.
How many functions did you find that failed to satisfy the conditions above? Did f (x) = x^2 work? What about f (x) =
x or f (x) = 3x + 7 or f (x) =
x
? Did you find an attribute common to those functions that did succeed? You should have, because there is only one extremely special family of functions that actually works here. Thus we return to our previous statement, in general, f (a + b) 6 = f (a) + f (b).
70 Relations and Functions
- For f (x) = 2 − x^2
- f (3) = − 7 • f (−1) = 1 • f
2
- f (4x) = 2 − 16 x^2 • 4 f (x) = 8 − 4 x^2 • f (−x) = 2 − x^2
- f (x − 4) = −x^2 + 8x − 14 • f (x) − 4 = −x^2 − 2 • f
x^2
= 2 − x^4
- For f (x) = x^2 − 3 x + 2
- f (3) = 2 • f (−1) = 6 • f
2
- f (4x) = 16x^2 − 12 x + 2 • 4 f (x) = 4x^2 − 12 x + 8 • f (−x) = x^2 + 3x + 2
- f (x − 4) = x^2 − 11 x + 30 • f (x) − 4 = x^2 − 3 x − 2 • f
x^2
= x^4 − 3 x^2 + 2
- For f (x) = (^) xx− 1
- f (3) = 32 • f (−1) = 12 • f
2
- f (4x) = (^4) x^4 x− 1 • 4 f (x) = (^) x^4 −x 1 • f (−x) = (^) x+1x
- f (x − 4) = x x−−^45 • f (x) − 4 = (^) x−x 1 − 4 = (^4) x−−^31 x - f
x^2
= x 2 x^2 − 1
- For f (x) = (^) x^23
- f (3) = 272 • f (−1) = − 2 • f
2
- f (4x) = (^321) x 3 • 4 f (x) = (^) x^83 • f (−x) = − (^) x^23
- f (x − 4) = (^) (x−^2 4) 3 = (^) x (^3) − 12 x (^22) +48x− 64 - f (x) − 4 = (^) x^23 − 4 = 2 −^4 x 3 x^3
- f
x^2
= (^) x^26
- For f (x) = 6
- f (3) = 6 • f (−1) = 6 • f
2
- f (4x) = 6 • 4 f (x) = 24 • f (−x) = 6
- f (x − 4) = 6 • f (x) − 4 = 2 • f
x^2
1.4 Function Notation 71
- For f (x) = 0
- f (3) = 0 • f (−1) = 0 • f
2
- f (4x) = 0 • 4 f (x) = 0 • f (−x) = 0
- f (x − 4) = 0 • f (x) − 4 = − 4 • f
x^2
- For f (x) = 2x − 5
- f (2) = − 1 • f (−2) = − 9 • f (2a) = 4a − 5
- 2 f (a) = 4a − 10 • f (a + 2) = 2a − 1 • f (a) + f (2) = 2a − 6
- f
a
= (^4) a − 5 = 4 −a^5 a
- f^ ( 2 a )= 2 a 2 − 5 • f (a + h) = 2a + 2h − 5
- For f (x) = 5 − 2 x
- f (2) = 1 • f (−2) = 9 • f (2a) = 5 − 4 a
- 2 f (a) = 10 − 4 a • f (a + 2) = 1 − 2 a • f (a) + f (2) = 6 − 2 a
- f
a
= 5 − (^4) a = 5 a a−^4
- f^ ( 2 a )= 5 − 22 a •^ f^ (a^ +^ h) = 5^ −^2 a^ −^2 h
- For f (x) = 2x^2 − 1
- f (2) = 7 • f (−2) = 7 • f (2a) = 8a^2 − 1
- 2 f (a) = 4a^2 − 2 • f (a + 2) = 2a^2 + 8a + 7 • f (a) + f (2) = 2a^2 + 6
- f
a
= (^) a^82 − 1 = 8 −a 2 a^2
(^2) − 1 2 •^ f^ (a^ +^ h) = 2a
(^2) + 4ah + 2 h^2 − 1
1.4 Function Notation 73
- For f (x) = (^2) x
- f (2) = 1 • f (−2) = − 1 • f (2a) = (^1) a
- 2 f (a) = (^4) a • f (a + 2) = (^) a+2^2 • f (a) + f (2) = (^2) a + 1 = a+2 2
- f
a
= a (^) • f^ ( 2 a )= (^1) a • f (a + h) = (^) a+^2 h
- For f (x) = 2x − 1, f (0) = −1 and f (x) = 0 when x = (^12)
- For f (x) = 3 − 25 x, f (0) = 3 and f (x) = 0 when x = (^152)
- For f (x) = 2x^2 − 6, f (0) = −6 and f (x) = 0 when x = ±
- For f (x) = x^2 − x − 12, f (0) = −12 and f (x) = 0 when x = −3 or x = 4
- For f (x) =
x + 4, f (0) = 2 and f (x) = 0 when x = − 4
- For f (x) =
1 − 2 x, f (0) = 1 and f (x) = 0 when x = (^12)
- For f (x) = (^4) −^3 x , f (0) = 34 and f (x) is never equal to 0
- For f (x) = 3 x (^2) − 12 x 4 −x^2 ,^ f^ (0) = 0 and^ f^ (x) = 0 when^ x^ = 0 or^ x^ = 4
- (a) f (−4) = 1 (b) f (−3) = 2 (c) f (3) = 0
(d) f (3.001) = 1. 999 (e) f (− 3 .001) = 1. 999 (f) f (2) =
- (a) f (4) = 4 (b) f (−3) = 9 (c) f (1) = 0
(d) f (0) = 1 (e) f (−1) = 1 (f) f (− 0 .999) ≈ 0. 0447
45. (−∞, 3] 46.
[
74 Relations and Functions
47. [− 3 , ∞) 48. (−∞, 7]
[ 1
3 ,^ ∞
3 ,^ ∞
[ 1
3 ,^3
[ 1
3 ,^6
55. (−∞, 8) ∪ (8, ∞) 56. [0, 8) ∪ (8, ∞)
57. (8, ∞) 58. [7, 9]
2 ,^ ∞
61. [0, 5) ∪ (5, ∞) 62. [0, 25) ∪ (25, ∞)
- A(3) = 9, so the area enclosed by a square with a side of length 3 inches is 9 square inches. The solutions to A(x) = 36 are x = ±6. Since x is restricted to x > 0, we only keep x = 6. This means for the area enclosed by the square to be 36 square inches, the length of the side needs to be 6 inches. Since x represents a length, x > 0.
- A(2) = 4π, so the area enclosed by a circle with radius 2 meters is 4π square meters. The solutions to A(r) = 16π are r = ±4. Since r is restricted to r > 0, we only keep r = 4. This means for the area enclosed by the circle to be 16π square meters, the radius needs to be 4 meters. Since r represents a radius (length), r > 0.
- V (5) = 125, so the volume enclosed by a cube with a side of length 5 centimeters is 125 cubic centimeters. The solution to V (x) = 27 is x = 3. This means for the volume enclosed by the cube to be 27 cubic centimeters, the length of the side needs to 3 centimeters. Since x represents a length, x > 0.
- V (3) = 36π, so the volume enclosed by a sphere with radius 3 feet is 36π cubic feet. The solution to V (r) = 323 π is r = 2. This means for the volume enclosed by the sphere to be 323 π cubic feet, the radius needs to 2 feet. Since r represents a radius (length), r > 0.
- h(0) = 64, so at the moment the object is dropped off the building, the object is 64 feet off of the ground. The solutions to h(t) = 0 are t = ±2. Since we restrict 0 ≤ t ≤ 2, we only keep t = 2. This means 2 seconds after the object is dropped off the building, it is 0 feet off the ground. Said differently, the object hits the ground after 2 seconds. The restriction 0 ≤ t ≤ 2 restricts the time to be between the moment the object is released and the moment it hits the ground.
- T (0) = 3, so at 6 AM (0 hours after 6 AM), it is 3◦^ Fahrenheit. T (6) = 33, so at noon ( hours after 6 AM), the temperature is 33◦^ Fahrenheit. T (12) = 27, so at 6 PM (12 hours after 6 AM), it is 27◦^ Fahrenheit.
