f a function.
SOLUTION
Notice thatf makes sense forx ∈ [−1,1](we may not take the square root of a negative number, so we cannot allowx >1 orx <−1). If we understand
f to have domain[−1,1]and rangeR, thenf : [−1,1] →Ris a function.
Math Note: When a function is given by a formula, as in Example 1.25, with no statement about the domain, then the domain is understood to be the set of allxfor which the formula makes sense.
You Try It: Let
g(x)= x
x2+4x+3.
What are the domain and range of this function?
EXAMPLE 1.26 Let f (x)= −3 ifx≤1 2x2 ifx >1 Determine whetherf isa function.
SOLUTION
Notice thatf unambiguously assigns toeachreal number another real num- ber. The rule is given in two pieces, but it is still a valid rule. Therefore it is a function with domain equal to Rand range equal toR. It is also perfectly correct to take the range to be(−4,∞), for example, sincefonly takes values in this set.
Math Note: One point that you should learn from this example is that a function may be specified bydifferent formulas on different parts of the domain.
Letf (x)= ± x. Discuss whetherf isa function.
SOLUTION
Thisf can only make sense forx ≥ 0. But even thenf isnota function since it is ambiguous. For instance, it assigns tox = 1 both the numbers 1 and−1.
1.8.2
GRAPHS OF FUNCTIONS
It is useful to be able to draw pictures which represent functions. These pictures, orgraphs, are a device for helping us to think about functions. In this book we will only graph functions whose domains and ranges are subsets of the real numbers.
We graph functions in thex-yplane. The elements of the domain of a function are thought of as points of thex-axis. The values of a function are measured on the
y-axis. The graph offassociates toxthe uniqueyvalue that the functionf assigns tox. In other words, a point(x, y)lies on the graph off if and only ify=f (x).
EXAMPLE 1.28
Let f (x) = (x2 +2)/(x−1). Determine whether there are pointsof the graph off corresponding tox =3,4, and 1.
SOLUTION
The y value corresponding to x = 3 isy = f (3) = 11/2. Therefore the point(3,11/2)lies on the graph off. Similarly,f (4)=6 so that(4,6)lies on the graph. However,f is undefined atx=1, so there is no point on the graph withx coordinate 1. The sketch in Fig. 1.38 was obtained by plotting several points.
Math Note: Notice that for eachxin the domain of the function there isone and only onepoint on the graph—namely the unique point withyvalue equal tof (x). Ifx is not in the domain off, then there is no point on the graph that corresponds tox.
EXAMPLE 1.29
CHAPTER 1
Basics
34
Fig. 1.38
Fig. 1.39
SOLUTION
Observe that, corresponding tox =3, for instance, there are twoy values on the curve. Therefore the curve cannot be the graph of a function.
You Try It: Graph the functiony=x+ |x|.
EXAMPLE 1.30
Fig. 1.40
SOLUTION
Notice that each x in the domain has just oney value corresponding to it. Thus, even though we cannot give a formula for the function, the curve is the graph of a function. The domain of this function is(−∞,3)∪(5,7).
Math Note: A nice, geometrical way to think about the condition that eachxin the domain has corresponding to it precisely oneyvalue is this:
If every vertical line drawn through a curve intersects that curve just once, then the curve isthe graph of a function.
You Try It: Use the vertical line test to determine whether the locusx2+y2=1 is the graph of a function.
1.8.3
PLOTTING THE GRAPH OF A FUNCTION
Until we learn some more sophisticated techniques, the basic method that we shall use for graphing functions is to plot points and then to connect them in a plausible manner.
EXAMPLE 1.31
Sketch the graph off (x)=x3−x.
SOLUTION
CHAPTER 1
Basics
36
x y=x3−x −3 −24 −2 −6 −1 0 0 0 1 0 2 6 3 24We plot these points on a pair of axes and connect them in a reasonable way (Fig. 1.41). Notice that the domain off is all ofR, so we extend the graph to the edges of the picture.
EXAMPLE 1.32
Sketch the graph of
f (x)=
−1 ifx ≤2 x ifx >2
SOLUTION
We again start with a table of values.
x y=f (x) −3 −1 −2 −1 −1 −1 0 −1 1 −1 2 −1 3 3 4 4 5 5
We plot these on a pair of axes (Fig. 1.42).
Since the definition of the function changes atx =2, we would be mistaken to connect these dots blindly. First notice that, for x ≤ 2, the function is identically constant. Its graph is a horizontal line. Forx >2, the function is a line of slope 1. Now we can sketch the graph accurately (Fig. 1.43).
Fig. 1.41
EXAMPLE 1.33
Sketch the graph off (x)=√x+1 .
SOLUTION
We begin by noticing that the domain off, that is the values ofxfor which the function makes sense, is {x:x ≥ −1}. The square root is understood to be the positive square root. Now we compute a table of values and plot some points.
CHAPTER 1
Basics
38
y x Fig. 1.42 y x Fig. 1.43 x y=√x+1 −1 0 0 1 1 √2 2 √3 3 2 4 √5 5 √6 6 √7Connecting the points in a plausible way gives a sketch for the graph off
(Fig. 1.44).
EXAMPLE 1.34
Sketch the graph ofx=y2.
TEAM
FLY
Fig. 1.44
SOLUTION
The sketch in Fig. 1.45 is obtained by plotting points. This curve isnotthe graph of a function.
Fig. 1.45
A curve that is the plot of an equation but which isnot necessarilythe graph of a function is sometimes called thelocusof the equation. When the curveis
the graph of a function we usually emphasize this fact by writing the equation in the formy =f (x).
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40
1.8.4
COMPOSITION OF FUNCTIONS
Suppose thatfandgare functions and that the domain ofgcontains the range off. This means that ifxis in the domain off thenf (x)makes sense but alsogmay be applied tof (x)(Fig. 1.46). The result of these two operations, one following the other, is calledgcomposedwithf or thecompositionofgwithf. We write
(g◦f )(x)=g(f (x)).
x f(x) g(f(x))
Fig. 1.46
EXAMPLE 1.35
Letf (x)=x2−1 andg(x)=3x+4. Calculateg◦f.
SOLUTION
We have
(g◦f )(x)=g(f (x))=g(x2−1). (∗)
Notice that we have started to workinside the parentheses: the first step was to substitute the definition off, namelyx2−1, into our equation.
Now the definition ofgsays that we takegofanyargument by multiplying that argument by 3 and then adding 4. In the present case we are applyinggto
x2−1. Therefore the right side of equation(∗)equals 3·(x2−1)+4.
This easily simplifies to 3x2+1. In conclusion,
g◦f (x)=3x2+1.
EXAMPLE 1.36
Letf (t)=(t2−2)/(t+1)andg(t)=2t+1. Calculateg◦f andf ◦g.
SOLUTION