Module 1 Lecture Notes

Home , Domain of a function, Range of a function

Contents 1.1 Identifying Functions...... 1 1.2 Algebraically Determining the Domain of a Function...... 4 1.3 Evaluating Functions...... 6 1.4 Function Operations...... 7 1.5 The Diﬀerence Quotient...... 9 1.6 Applications of Function Operations...... 10 1.7 Determining the Domain and Range of a Function Graphically.... 12 1.8 Reading the Graph of a Function...... 14

1.1 Identifying Functions

In previous classes, you should have studied a variety basic functions. For example, 1 √ f(x) = 3x − 5, g(x) = 2x2 − 1, h(x) = , j(x) = 5x + 2 x − 5 We will begin this course by studying functions and their properties. As the course progresses, we will study inverse, composite, exponential, logarithmic, polynomial and rational functions. Math 111 Module 1 Lecture Notes

Definition 1: A relation is a correspondence between two variables. A relation can be ex- pressed through a set of ordered pairs, a graph, a table, or an equation. A set containing ordered pairs (x, y) defines y as a function of x if and only if no two ordered pairs in the set have the same x-coordinate. In other words, every input maps to exactly one output. We write y = f(x) and say “y is a function of x.” For the function defined by y = f(x), • x is the independent variable (also known as the input)

• y is the dependent variable (also known as the output)

• f is the function name

Example 1: Determine whether or not each of the following represents a function.

Table 1.1 Chicken Name Egg Color Emma Turquoise Hazel Light Brown George(ia) Chocolate Brown Isabella White Yvonne Light Brown

(a) The set of ordered pairs of the form (chicken name, egg color) shown in Table 1.1. This deﬁnes a function. For each input, there is exactly one output. More speciﬁcally, if you are told the name of the chicken, you know for certain what the egg color will be.

(b) The set of ordered pairs of the form (egg color, chicken name) shown in Table 1.1. This DOES NOT deﬁne a function. For every input, there is not exactly one output. More speciﬁcally, if you are told the color is Light Brown, you have no way of knowing if the egg belongs to Hazel or Yvonne.

Example 2: In Table 1.2, determine for which relations y is a function of x. Notice in Table 1.2 that there are a lot of similarities between the equations x2 + y = 4 and x + y2 = 4. However, we are determining if y is a function of x and thus are determining if every x-value is associated with exactly one y-value—not the reverse.

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Table 1.2: Identifying Functions Relation Function? Justiﬁcation

x y 1 1 No The input 2 has more than one output. 2 1 2 3

x y 8 49 Yes Every input has exactly one output. 12 60 16 49

x + y = 4 Yes There is exactly one y-value associated with every x-value. Note that the equation can be written explicitly in terms of y: y = 4 − x.

x2 + y = 4 Yes There is exactly one y-value associated with every x-value. Note that the equation can be written explicitly in terms of y: y = 4 − x2.

2 x + y = 4 No The outputs 2 and −2 both√ correspond to the input 0. Furthermore, y = ± 4 − x and y cannot be written explicitly in terms of x.

2 2 x + y = 25 No The outputs 3 and −3 both√ correspond to the input 4. Furthermore, y = ± 25 − x2 and y cannot be written explicitly in terms of x.

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1.2 Algebraically Determining the Domain of a Function

Deﬁnition 2: Domain and Range • The domain of a function is the set of all possible inputs.

• The range of a function is the set of all possible outputs.

The largest possible set for each the domain and the range is the set of all real numbers. So far, there are two possible circumstances that restrict the domain of a function. When a number causes either of the following to occur, it must be excluded from the domain of a function: • Division by zero

• The square root (or any even root) of a negative number

Example 3: State the domain of each of the following functions in both set-builder and interval notation. √ (a) f(x) = 7 − 2x Since this function contains a radical, we know that 7 − 2x ≥ 0. Solving for x, we ﬁnd:

7 − 2x ≥ 0 7 ≥ 2x 7 ≥ x 2 7 x ≤ 2

7 7 The domain of f is x | x ≤ 2 . In interval notation, we write −∞, 2

(b) s(t) = 5t2 + 4t + 3 Noting that s is a quadratic function (and does not have division by zero or contain a radical), we know that the domain of s is all real numbers. In set-builder notation, we write {t | t is a real number}. In interval notation, we write (−∞, ∞).

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x2 − 7x + 10 (c) g(x) = x2 + x − 6 Since this function contains x2 + x − 6 in the denominator, we know that we must exclude values of x such that x2 + x − 6 = 0 from the domain:

x2 + x − 6 = 0 (x + 3)(x − 2) = 0 x + 3 = 0 or x − 2 = 0 x = −3 or x = 2

The domain of f is {x | x 6= −3, 2}. In interval notation, we write (−∞, −3) S (−3, 2) S (2, ∞).

x + 2 (d) m(x) = x2 − 1 Since this function contains x2 − 1 in the denominator, we know that we must exclude values of x such that x2 − 1 = 0 from the domain:

x2 − 1 = 0 x2 = 1 √ x = ± 1 x = ±1

The domain of m is {x | x 6= −1, 1}. In interval notation, we write (−∞, −1) S (−1, 1) S (1, ∞).

√ 3 t − 2 (e) k(t) = 5t + 40 Since this function contains 5t + 40 in the denominator, we know that we must exclude values of t such that 5t + 40 = 0 from the domain:

5t + 40 = 0 5t = −40 t = −8

The domain of k is {t | t 6= −8}. In interval notation, we write (−∞, −8) S (−8, ∞). Note that the radical with an odd power does NOT limit the domain in any way.

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4t + 1 (f) g(t) = √ 5 − t √ Since this function√ contains 5 − t in the denominator, we know that we must exclude values of t such that 5 − t = 0. Furthermore, since 5 − t is inside the radical, we must have 5 − t ≥ 0. Combined, this means that 5 − t > 0. Solving this for t: 5 − t > 0 5 > t t < 5 The domain of g is {t | t < 5}. In interval notation, we write (−∞, 5).

√ x − 5 (g) q(x) = x − 9 Since the function q contains x − 9 in the denominator, we know that values of x such that x − 9 = 0 must be excluded from the domain. Thus x 6= 9. Furthermore, since the function also has x − 5 inside a radical, we know that x − 5 ≥ 0. Thus x ≥ 5. Combined, we have x ≥ 5 and x 6= 9. The domain of q is then {x | x ≥ 5 and x 6= 9}. In interval notation, we write [5, 9) S (9, ∞).

1.3 Evaluating Functions

Recall that to evaluate a function is to ﬁnd the function’s value at a given input. One of the primary motivations of using function notation is that it is a means of communicating both the input and the output. For example, if we deﬁne a function by f(x) = x2, when we evaluate the function at x = −3, we see f(−3) = (−3)2 = 9 The statement f(−3) = 9 communicates both the input and the output. Conversely, if we had let y = x2 and found the value of y when x = −3, we would have seen y = (−3)2 = 9 Our result is the same: y = 9. But this does not tell us what the original input was!

x2 Example 4: Let f(x) = . Evaluate the following, simplifying each expression as much as x + 1 possible:

(a) f(−3) (c) f(0) (e) f(−x) (g) f(2x) (i) f(x + h) (b) f(1) (d) f(t) (f) −f(x) (h) f(x − 2)

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(a) (f) (−3)2 f(−3) = x2 −3 + 1 −f(x) = − 9 x + 1 = − 2 (g) (b) (1)2 (2x)2 f(1) = f(2x) = 1 + 1 2x + 1 1 4x2 = = 2 2x + 1 (c) (0)2 (h) f(0) = 0 + 1 (x − 2)2 = 0 f(x − 2) = (x − 2) + 1 (d) x2 − 4x + 4 = x − 1 (t)2 f(t) = t + 1 (i) (e) (−x)2 (x + h)2 f(−x) = f(x + h) = −x + 1 (x + h) + 1 x2 x2 + 2xh + h2 = = −x + 1 x + h + 1

1.4 Function Operations

Function Operations • The sum of f and g, f + g, is deﬁned by (f + g)(x) = f(x) + g(x).

• The diﬀerence of f and g, f − g, is deﬁned by (f − g)(x) = f(x) − g(x).

• The product of f and g, f · g, is deﬁned by (f · g)(x) = f(x) · g(x).

f f f(x) • The quotient of f and g, g , is deﬁned by g (x) = g(x) .

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x 3 Example 5: Let f(x) = and let g(x) = . Find and fully simplify (f +g)(x), (f ·g)(x), x + 3 x + 3 g and (x). State the domain of f + g, f · g, and g . f f

(f + g)(x) = f(x) + g(x) x 3 = + x + 3 x + 3 x + 3 = x + 3 = 1, x 6= −3 The domain of f + g is {x | x 6= −3}. This could also be written in interval notation as (−∞, −3) S(−3, ∞).

(f · g)(x) = f(x) · g(x) 3 x = · x + 3 x + 3 3x = (x + 3)2 The domain of f · g is {x | x 6= −3}. This could also be written in interval notation as (−∞, −3) S(−3, ∞).

g g(x) (x) = f f(x) 3 x + 3 = x x + 3 3 x = ÷ x + 3 x + 3 3 x + 3 = · x + 3 x 3 = , x 6= −3 x g The domain of f is {x | x 6= −3, 0}. This could also be written in interval notation as (−∞, −3) S(−3, 0) S(0, ∞).

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1.5 The Diﬀerence Quotient

f(x + h) − f(x) In calculus, the expression , h 6= 0 is used to ﬁnd the average rate of change of a h function. This expression is known as the diﬀerence quotient.

Example 6: Calculate the diﬀerence quotient for the function deﬁned by f(x) = 7x + 4.

f(x + h) − f(x) (7(x + h) + 4) − (7x + 4) = h h 7x + 7h + 4 − 7x − 4 = h 7h = 7 = 7, h 6= 0

Example 7: Find and completely simplify the diﬀerence quotient for f(x) = 3x2 − 9x + 2.

f(x + h) − f(x) (3(x + h)2 − 9(x + h) + 2) − (3x2 − 9x + 2) = h h (3(x2 + 2xh + h2) − 9(x + h) + 2) − (3x2 − 9x + 2) = h 3x2 + 6xh + 3h2 − 9x − 9h + 2 − 3x2 + 9x − 2 = h 6xh + 3h2 − 9h = h h(6x + 3h − 9) = h = 6x + 3h − 9, h 6= 0

Example 8: Find and completely simplify the diﬀerence quotient for f(x) = −5. This problem can seem a lot trickier than necessary. Since f(x) = −5, f(1) = −5, f(2) = −5, f(10) = −5, f(a) = −5, etc. Similarly, f(x + h) = −5. Thus: f(x + h) − f(x) −5 − (−5) = h h 0 = h = 0, h 6= 0

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1.6 Applications of Function Operations

Example 9: Let P (t) be the population of a country (in millions) t years after 2000, where P (t) = 0.01t2 + 2. Let N(t) be the number of people (in millions) that a country can feed t years after 2000, where N(t) = 4 + 0.5t. Use this to complete the following problems. (a) One way of measuring prosperity is to measure the surplus. The surplus, S(t), is deﬁned to be N(t) − P (t). Compute this function. (b) Evaluate and interpret S(45). (c) Evaluate and interpret S(55). (d) Another way of measuring prosperity is to calculate the per capita food supply, R(t), which N(t) is deﬁned by . Compute this function. P (t) (e) Evaluate and interpret R(45). (f) Evaluate and interpret R(55). (g) Graph these two functions and interpret their point of intersection. (a)

S(t) = N(t) − P (t) = (4 + 0.5t) − (0.01t2 + 2) = −0.01t2 + 0.5t + 2

(b)

S(45) = −0.01(45)2 + 0.5(45) + 2 = 4.25

According to this model, in 2045 the country will be able to feed 4.25 million more people than it needs to.

(c)

S(55) = −0.01(55)2 + 0.5(55) + 2 = −0.75

According to this model, in 2055 the country will need to increase its food production so that it can feed 750,000 more people.

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(d) N(t) R(t) = P (t) 4 + 0.5t = 0.01t2 + 2

(e) 4 + 0.5(45) R(45) = 0.01(45)2 + 2 ≈ 1.2 According to this model, in 2045 the country will be able to feed about 1.2 times as many people as it needs to. (f) 4 + 0.5t R(55) = 0.01t2 + 2 ≈ 0.98 According to this model, in 2055 the country will be able to feed about 0.98 times as many people as it needs to (or 98% of its population). (g) As shown in Figure 1.1, the point of intersection is approximately (53.72, 30.86). (This can be found using the intersection command on a graphing calculator.) The point of intersection represents when the amount of food the country is producing is the same as the amount it needs to feed its population.

Figure 1.1

y = P (t) y = N(t)

40 , millions of people y

20 40 60 80 t, years after January 1, 2000

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1.7 Determining the Domain and Range of a Function Graphically

Recall that the domain of a function is the set of all possible inputs, while the range of a function is the set of all possible outputs. To graphically determine the domain, scroll along the horizontal axis. Are there places where the function is not deﬁned? These numbers must be excluded from the domain. Are there endpoints or arrows on the end of the graph? If there are endpoints, recall that open circles show that the endpoint is NOT included, while closed circles show that the endpoint is included. Arrows show that the function continues in that manner indeﬁnitely. Once you’ve determined the domain, use the same process on the vertical axis to determine the range. Note: You can state the domain and range for a graph that represents a relation in the exact same way you do for a function.

Example 10: Determine if the graphs in Figures 1.2-1.5 are functions. State the domain and range for each graph using interval notation.

Figure 1.2 Figure 1.3 y y

6 6

4 4

2 2 x x

−6 −4 −2 2 4 6 −6 −4 −2 2 4 6

−2 −2

−4 −4

−6 −6

Function? NO Function? YES Domain: (−∞, −1] Domain: (1, 6] Range: (−∞, ∞) Range: (−7, 7]

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Figure 1.4 Figure 1.5 y y

6 6

4 4

2 2 x x

−6 −4 −2 2 4 6 −6 −4 −2 2 4 6

−2 −2

−4 −4

−6 −6

Function? YES Function? YES Domain: (−∞, 1) S[3, ∞) Domain: {−6, −4, −2, 1, 2, 3, 4, 6} Range: (−∞, −4] S(2, ∞) Range: {−7, −5, −2, 1, 2, 3, 5, 7}

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1.8 Reading the Graph of a Function

Example 11: Use the graph of y = f(x) in Figure 1.6 to answer the following.

Figure 1.6: Graph of y = f(x) y 6

2 x

−10 −8 −6 −4 −2 2 4 6 8 10 −2

−4

−6

(a) Find f(0) and f(−2). (g) What is the vertical intercept? f(0) = −6 and f(−2) = −5 The vertical intercept is (0, −6).

(b) Evaluate f at 4. (h) Solve f(x) = −4. State the solution set. The value of f at 4 is −1. {x | x = −7, −4, 3}

(d) For what values of x is f(x) = 0? (j) For what values of x is f(x) > 0? f(x) = 0 when x = −9 or x = 7 −10 < x < −9 or 7 < x ≤ 8

(e) State the zeros of f. (k) State the domain of f. The zeros of f are −9 and 7. (−10, 8]

(f) What are the horizontal intercepts? (l) State the range of f. The horizontal intercepts are (−9, 0) and [−7, 5) (7, 0).

Example 12: Bees! A population of bees was happily residing in someone’s backyard last year. Let B(t) the size of the bee population (in thousands) t months after April 1, 2012. This function is modeled in Figure 1.7.

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Figure 1.7: Graph of y = B(t)

4 , thousands of bees y 2

2 4 6 8 10 t, months after April 1, 2012

(a) Find B(3). Explain what this function value represents in the context of the problem. B(3) = 15 Three months after April 1, 2012 (or on July 1, 2012), there were 15,000 bees.

(b) Find B(0). Explain what this function value represents in the context of the problem. B(0) = 10 On April 1, 2012, there were 10,000 bees.

B(t) = 13 t = 1 or 5 Solution set: {1, 5} There were 13,000 bees on May 1, 2012 and again on September 1, 2012.

(d) Solve B(t) = 3. Explain what this solution set represents in the context of the problem.

B(t) = 3 t = 8 Solution set: {8} There were 3,000 bees 8 months after April 1, 2012 (or on December 1, 2012).

(e) State the domain and range of B. Domain: {t | 0 ≤ t ≤ 9} (Set-builder Notation) [0, 9] (Interval Notation) Range: {y | 0 ≤ y ≤ 15} (Set-builder notation) [0, 15] (Interval Notation)

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