In other words, if the input is positive or nonnegative, then the output is identical to the input; if the input is not positive, then change its sign to obtain the 8 We have already dis
Trang 110 Which of the following functions is continuous at x = 2?
(a) f (x) = x + 3
(b) f (x) =x+3x−2
(c) f (x) =x2+x−6x−2
11 Which of the following functions is continuous at x = 1?
(a) f (x) = 5
(b) f (x) =5xx
(c) f (x) =5x(x−1)x−1
(d) f (x) =x2x−1(x−1)
2.2 A POCKETFUL OF FUNCTIONS: SOME BASIC EXAMPLES
Throughout this text we will be looking at functions analytically and graphically Consider any equation relating two variables Every point that lies on the graph of the equation satisfies the equation Conversely, every point whose coordinates satisfy the equation lies
on the graph of the equation The correspondence between pairs of x and y that constitute the coordinates of a point on a graph and that satisfy an equation was a key insight of Rene Descartes (1596–1650) that unified geometry and algebra.8As a consequence of this insight, the fields of geometry and algebra became irrevocably intertwined, opening the door to much of modern mathematics, including calculus
We have been discussing functions and their graphs in a very general way, but it is always strategically wise to include a few simple, concrete examples for reference In this section you will become familiar with a small sampling of functions, functions that you can carry about and pull out of your back pocket at any moment As we go along in the text this collection will grow; you will learn that these functions belong to larger families of functions sharing some common characteristics, and you’ll also be introduced to a greater variety of families of functions But for the time being, we will become familiar with a few individual functions
A Few Basic Examples
Consider the following function:
f (x) = x g(x) = x2 h(x) = |x| j (x) =1x
The function f is the identity function; its output is identical to its input.
The function g is the squaring function; its output is obtained by squaring its input The function h is the absolute value function; its output is the magnitude (size) of
its input In other words, if the input is positive (or nonnegative), then the output is identical to the input; if the input is not positive, then change its sign to obtain the
8 We have already discussed this correspondence when looking at the graph of a function In fact, it holds for the graph of any
Trang 2output We can define the absolute value function |x| analytically as follows.
|x| =
x if x ≥ 0;
−x if x < 0
The function j is the reciprocal function; the output is the reciprocal of the input.
output =input1 The graphs of these functions are given below Familiarize yourself with these functions by doing the exercises that follow
x g
g (x) = x 2
f (x) = x h (x) = |x| j (x) =
1
1
x f
1
1
x h
1
1
x j
1 1
–1
x
Figure 2.11
EXERCISE 2.5 Look at each of the functions f (x) = x, g(x) = x2, h(x) = |x|, and j (x) =1x one by one
and answer the following questions
(a) What are the domain and range of the function?
(b) Where is the function positive? Negative?
(c) Where is the function increasing? Decreasing?
(d) Is the function continuous? If it is not continuous everywhere, where is it discontinuous? (e) Is the function 1-to-1?
Answers are provided at the end of the section.
EXERCISE 2.6 Consider the function j (x) = 1x
(a) Why is j (x) never equal to zero?
(b) Why is zero not in the domain of j ? (c) As x increases without bound what happens to the values of j ? As x decreases (becomes more and more negative) without bound what happens to the values of j ? How is this information displayed on the graph of j ? How is it displayed on your graphing calculator’s rendition of the graph?
(d) Find j (x) for x = 0.01, 0.001, and 0.0001 If x is a positive number, as x gets increasingly close to zero, what happens to the values of j ? How is this information displayed on the graph?
(e) Find j (x) for x = −0.01, −0.001, and −0.0001 If x is a negative number, as x gets increasingly close to zero, what happens to the values of j ? How is this information
Trang 3displayed on the graph? How is this displayed on your calculator’s rendition of the graph?
(f ) Fill in the blanks with >, <, or = as appropriate
i If x > 0 then j (x) 0
ii If x < 0 then j (x) 0
iii If |x| > 1 then |j (x)| 1
iv If 0 < |x| < 1 then |j (x)| 1
(g) Translate the statements from part (f ) into plain prose Your translation should be accessible to someone who knows nothing about functions and absolute values
EXERCISE 2.7 Consider the function g(x) = x2
(a) Fill in the blanks with >, <, or = as appropriate
i If |x| 1, then g(x) > |x|
ii If |x| 1, then g(x) < |x|
(b) Translate the statements from part (a) into plain prose Your translation should be accessible to someone who knows nothing about functions and absolute values (c) Consider h(x) = |x|
i For what values of x is h(x) > g(x)?
ii For what values of x is h(x) < g(x)?
REMARKS
The graph of f (x) = x given in Figure 2.11 has the same scale on the x- and y-axes and therefore the line cuts the angle between the axes evenly in two If different scales were used on the axes this would not be true.9
For x ≥ 0 the functions f (x) = x and h(x) = |x| are identical The graph of h(x) has
a sharp corner at x = 0 No matter how much you magnify the portion of the graph around that point, the corner remains sharp
The graph of the squaring function g(x) = x2is called a parabola Notice that g(−3) = (−3)2, not −32 Therefore g(3) = g(−3) and, more generally, g(x) = g(−x) The
graph of g is symmetric about the y-axis; the graph is a mirror image on either side of
the y-axis You could fold the graph along the y-axis and the function would exactly match itself on either side of the y-axis The characteristic g(x) = g(−x) guarantees this symmetry
The graph of j (x) =x1has a vertical asymptote at x = 0 and a horizontal asymptote at
y = 0
If y =1x, then y is inversely proportional to x
9 If you use a graphing calculator to look at the graph of a function you can set the range and domain so that the scales are very different You should be aware that your choice of domain and range can make the line f (x) = x look almost horizontal or
Trang 4D e f i n i t i o n
Ais inversely proportional to B if
A = k B for some constant k As B increases A decreases, and vice versa
This type of relationship arises frequently in the world around us For example, the amount of time required to complete a trip is inversely proportional to the rate at which one travels
Informal Introduction to Asymptotes The function Q(x) has a vertical asymptote at x = a if Q is undefined at x = a and as x
gets closer and closer to a the magnitude of Q(x), written |Q(x)|, increases without bound Because Q is undefined at its vertical asymptotes, the graph of a function will never cross its vertical asymptote Around a vertical asymptote, the graph will resemble one of the four options shown below If Q has a one-sided vertical asymptote at x = a then as x approaches
afrom one side |Q(x)| increases without bound
Figure 2.12
For the function j (x) = 1x, as x approaches 0 from the right, the value j (x) grows without bound We can write this symbolically:
as x → 0+ j (x) → ∞
“As x → 0+” means “as x approaches zero from the right”; “j (x) → ∞” means “j (x) grows without bound.” On the other hand, as x approaches 0 from the left, the value of j (x) gets increasingly negative without bound We can write this symbolically:
as x → 0− j (x) → −∞
The function Q has a horizontal asymptote at y = c if, as the magnitude of x increases
without bound, the value of Q(x) gets closer and closer to c A horizontal asymptote tells us only about the behavior of the function for |x| very large; therefore, it can be crossed A one-sided horizontal asymptote tells us what happens to Q as x either increases or decreases
Trang 5without bound It can be crossed As |x| increases without bound, j (x) = 1x approaches zero We can express this symbolically by writing
as x → ∞ j (x) → 0 and as x → −∞ j (x) → 0
Even and Odd Symmetry
A function may be even, odd, or neither
D e f i n i t i o n
A function f is even if f (−x) = f (x) for all x in the domain of f This means
that the height of the graph of f is the same at x and at −x The graph of an even function is a mirror image about the y-axis
A function is odd if f (−x) = −f (x) for all x in the domain of f An odd function
is said to be symmetric about the origin For further description see the answers
to Exercise 2.8
x f
a
f
a -a x
f
x
f
(-a, f(-a)) (a, f(a))
even
f(x) = f(-x)
f(x) = -f(-x) (-a, f(-a))
(a, f(a))
Figure 2.13
EXERCISE 2.8 In this exercise we return to the four functions we introduced at the beginning of this section:
f (x) = x, g(x) = x2, h(x) = |x|, and j (x) =x1 (a) Which of functions f , g, h, and j are even?
(b) Which of functions f , g, h, and j are odd?
(c) Describe in words the characteristic of the graph of an odd function (The graph of an
odd function is not a mirror about the x-axis.)
Answers to Exercise 2.8 are provided at the end of the section.
Working with Absolute Values
To look at the absolute value of a quantity is to look only at its magnitude, to make it nonnegative
Trang 6Analytic Principle for Working with Absolute Values
To deal with an absolute value, remove it by breaking the problem up into two cases.
Case (1): If the expression inside the absolute value is nonnegative, then simply remove the absolute value signs
Case (2): If the expression inside the absolute value is negative, then change its sign and remove the absolute values It follows that:
|x − 3| = x − 3 for x ≥ 3,
−(x − 3) for x ≤ 3
Geometric Principle for Working with Absolute Values
Equivalent to the preceding analytic definition is the “geometric” definition: The absolute value of x is the distance between x and 0 on the number line More generally, this geometric definition tells us that |x − a| is the distance between x and a on the number line It follows that |x − 3| is the distance between x and 3, while |x + 2| (which equals |x − (−2)|) is the distance between x and −2
We can use this geometric interpretation to solve equations and inequalities involving absolute values For example, we can interpret the inequality
|x − 100| ≤ 0.1
as “x differs from 100 by no more than 0.1,” or x ∈ [99.9, 100.1]
(a) |x − 1| = 2 (b) |x + 1| ≤ 2 (c) |2x + 3| > 1 SOLUTIONS
(a) Geometric Approach:
|x − 1| = 2 (the distance between x and 1) = 2
Figure 2.14
x = −1 or x = 3
Trang 7Analytic Approach:
Case (1): x − 1 ≥ 0 Case (2): x − 1 < 0
x = −1
Check that when x = 3, x − 1 ≥ 0 and when x = −1, x − 1 < 0 Alternatively, check the answers in the original problem
(b) Geometric Approach:
|x + 1| ≤ 2
|x − (−1)| ≤ 2 (the distance between x and −1) ≤ 2
– 4 –3 –2 –1 0 1 2
3
Figure 2.15
xis between −3 and 1, including both endpoints
x ∈ [−3, 1]
Analytic Approach: Solve the corresponding equation: |x + 1| = 2
Case (1): x + 1 ≥ 0 Case (2): x + 1 < 0
x = −3 These two solutions partition the number line into three intervals To determine whether each of these intervals contains acceptable values of x, substitute one number from each
of these intervals into the original equation This technique gives us −3 ≤ x ≤ 1 (c) Geometric Approach:
|2x + 3| > 1
|2x − (−3)| > 1 (the distance between 2x and −3) > 1 We’ll focus on 2x and solve for x later
Trang 81 1
– 4 –3 –2 –1 0 1 –6 –5
Figure 2.16
2x < −4 or 2x > −2
x < −2 or x > −1 Analytic Approach: Solve the corresponding equation |2x + 3| = 1
Case (1):2x + 3 ≥ 0 Case (2):2x + 3 < 0
x = −2 These two solutions once again divide the number line into three intervals Check to see if a number inside each of these intervals is a solution to determine whether all numbers within the interval are solutions Here we discover x < −2 or x > −1
EXERCISE 2.9 Solve for x Do these problems in two ways, first using the analytic definition of the absolute
value and then using the geometric definition of the absolute value
(a) |3 − x| = 2 (b) |2x + 1| = 4
Answers to Exercise 2.9 are provided at the end of the section.
Answers to Selected Exercises
Answers to Exercise 2.5
The linear function f (x) = x (a) The domain of f is all real numbers Likewise, the range of f is all real numbers (b) f is positive for x positive, negative for x negative
(c) The graph of f is a straight line The graph is always increasing
(d) f is a continuous function
(e) f is 1-to-1
g(x) = x2 (a) The domain of g is all real numbers The range of g is all nonnegative real numbers (b) g is positive for all x except x = 0 g(0) = 0
(c) g is decreasing on (−∞, 0] and increasing on [0, ∞)
(d) g is a continuous function
Trang 9(e) g is not 1-to-1 By restricting the domain to nonnegative real numbers the function
can be made 1-to-1
The absolute value function h(x) = |x|
(a) The domain of h is all real numbers The range of h is all nonnegative real numbers (b) h is positive for all x except x = 0 h(0) = 0
(c) h is decreasing on (−∞, 0] and increasing on [0, ∞)
(d) h is a continuous function
(e) h is not 1-to-1 By restricting the domain to nonnegative real numbers the function
can be made 1-to-1
j (x) =1x
(a) The domain of j is all nonzero real numbers The range of j is all nonzero real numbers
(b) j is positive for x positive, negative for x negative
(c) j is decreasing on (−∞, 0) and decreasing on (0, ∞)
(d) j is undefined and discontinuous at x = 0
(e) j is 1-to-1
Answers to Exercise 2.8
(a) g and h are even functions
(b) f and j are odd functions
(c) The graph of an odd function is said to have symmetry about the origin What does this mean? Graph the function for all nonnegative x in the domain The other half of the graph can be obtained by reflecting this portion of the graph across the y-axis and then across the x-axis (This is equivalent to rotating the selected portion of the graph 180◦
around the origin.)
Answers to Exercise 2.9
(a) i Analytic Approach: Remove the absolute value signs by breaking the problem into two cases—one in which the expression inside the absolute value is positive and the other in which the expression inside the absolute value sign is not positive Case (1): The expression inside the absolute value is positive Because 3 − x > 0,
we replace |3 − x| by 3 − x
3 − x = 2
−x = −1
x = 1 Case (2): The expression inside the absolute value is not positive Because
3 − x ≤ 0, we replace |3 − x| by −(3 − x) = −3 + x
−3 + x = 2
x = 5 Check that these answers satisfy the original equation
ii Geometric Approach:
|3 − x| = 2 (the distance between 3 and x) = 2
Trang 100 1 2 3 4 5 6
x = 1 or x = 5
Figure 2.17
(b) i Analytic Approach: Remove the absolute value signs by breaking the problem down into two cases
Case (1): The expression inside the absolute value is positive Because 2x + 1 > 0,
we replace |2x + 1| by 2x + 1
2x + 1 = 4 2x = 3
x = 3 2 Case (2): The expression inside the absolute value is not positive
Because 2x + 1 ≤ 0, we replace |2x + 1| by −(2x + 1) = −2x − 1
−2x − 1 = 4
−2x = 5
x = −5 2 Check that both of these answers satisfy the original equation
ii Geometric Approach:
|2x + 1| = 4
|2x − (−1)| = 4 (the distance between 2x and − 1) = 4
0 1 2 3 4 –6 –5 – 4 –3 –2 –1
Figure 2.18
2x = −5 or 2x = 3
x = −5
2
P R O B L E M S F O R S E C T I O N 2 2
1 This problem applies the definitions of proportionality (also referred to as direct pro-portionality), and inverse proportionality