Chapter 1
Functions
In this chapter we review the basic concepts of functions, polynomial functions, rational functions, trigonometric functions, logarithmic functions, exponential functions, hyperbolic functions, algebra of functions, composition of functions and inverses of functions.
1.1 The Concept of a Function
Basically, a functionf relates each element x of a set, say Df , with exactly one element y of another set, say Rf . We say that Df is the domain of f and Rfis therangeof f and express the relationship by the equation y = f(x).It is customary to say that the symbol x is an independent variable and the symbol y is the dependent variable.
Example 1.1.1 Let Df = {a, b, c}, Rf = {1, 2, 3} and f(a) = 1, f(b) = 2 and f(c) = 3. Sketch the graph of f.
Example 1.1.2 Sketch the graph of f(x) = |x|.
Let Df be the set of all real numbers and Rf be the set of all non-negative real numbers. For each x in Df , let y = |x| in Rf . In this case, f(x) = |x|,
|
the absolute value of x. Recall that |
We note that f(0) = 0, f(1) = 1 and f(minus;1) = 1.
If the domain Df and the range Rf of a function f are both subsets of the set of all real numbers, then the graphof f is the set of all ordered pairs (x, f(x)) such that x is in Df . This graph may be sketched in the xy-coordinate plane, using y = f(x). The graph of the absolute value function in Example 2 is sketched as follows:
Example 1.1.3 Sketch the graph of
f(x)=
In order that the range of f contain real numbers only, we must impose the restriction that xge; 4. Thus, the domain Df contains the set of all real numbers x such that xge; 4. The range Rf will consist of all real numbers y such that yge; 0. The graph off is sketched below.
Example 1.1.4 A useful function in engineering is the unit step function, u, defined as follows:
The graph of u(x) has an upward jump at x = 0. Its graph is given below.
Example 1.1.5 Sketch the graph of
f(x)=
It is clear that Df consists of all real numbers xne;plusmn;2. The graph of f is given below.
We observe several things about the graph of this function. First of all, the graph has three distinct pieces, separated by the dotted vertical lines x = minus;2 and x = 2. These vertical lines, x = plusmn;2, are called theverticalasymptotes. Secondly, for large positive and negative values ofx, f(x) tendsto zero. For this reason, the x-axis, with equation y = 0, is called a horizontalasymptote.
Let f be a function whose domain Df and range Rf are sets of real numbers. Then f is said to be even if f(x) = f(minus;x) for all x in Df . Andf is said to be odd if f(minus;x) = minus;f(x) for all x in Df . Also, f is said to be one-to-one iff(x1) =f(x2) implies thatx1=x2.
Example 1.1.6 Sketch the graph of f(x) = x4minus;x2. This function f is even because for all x we have
f(minus;x) = (minus;x)4 minus; (minus;x)2= x4 minus; x2= f(x).
The graph of f is symmetric to the y-axis because (x, f(x)) and (minus;x, f(x)) are on the graph for every x. The graph of an even function is always symmetricto the y-axis. The graph offis given below.
This function f is not one-to-one because f(minus;1) = f(1).
Example 1.1.7 Sketch the graph of g(x) = x3minus; 3x.
The function g is an odd function because for each x,
g(minus;x) = (minus;x)3 minus; 3(minus;x) = minus;x3 3x = minus;(x3 minus; 3x) = minus;g(x).
The graph of this function g is symmetric to the origin because (x, g(x)) and (minus;x,minus;g(x)) are on the graph for all x. The graph of an odd function isalways symmetric to the origin. The graph ofgis given below.
|
This function gis not one-to-one because g(0)=g()=g(﹣) |
It can be shown that every function f can be written as the sum of an even function and an odd function. Let
g(x) =(f(x) f(minus;x)), h(x) =(f(x) minus; f(minus;x)).
Then,
g(minus;x) =(f(minus;x) f(x)) = g(x)
h(minus;x) =(f(minus;x) minus; f(x)) = minus;h(x).
Furthermore
f(x) = g(x) h(x).
Example 1.1.8 Express f as the sum of an even function and an odd function, where,
f(x) = x4 minus; 2x3 x2 minus; 5x 7.
We define
g(x) =(f(x) f(minus;x))
{(x4 minus; 2x3 x2 minus; 5x 7) (x4 2x3 x2 5x 7)}
=x4 x2 7
and
h(x) =(f(x) minus; f(minus;x))
{(x4 minus; 2x3 x2 minus; 5x 7) minus; (x4 2x3 x2 5x 7)}
=minus;2x3 minus; 5x.
Then clearly g(x) is even and h(x) is odd.
g(minus;x) = (minus;x)4 (minus;x)2 7
=x4 x2 7
=g(x)
h(minus;x) = minus; 2(minus;x)3 minus; 5(minus;x)
=2x3 5x
=minus;h(x).
We note that
g(x) h(x) = (x4 x2 7) (minus;2x3 minus; 5x)
=x4 minus; 2x3 x2 minus; 5x 7
=f(x).
It is not always easy to tell whether a function is one-to-one. The graphical test is that if no horizontal line crosses the graph of f more than once, then f is one-to-one. To show that f is one-to-one mathematically, we need to show that f(x1) = f(x2) implies x1 = x2.
Example 1.1.9 Show that f(x) = x3
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第一章 函数
在本章节中我们复习回顾函数的基本概念，多项式函数，有理函数，三角函数，幂函数，双曲函数，代数函数，函数的组合和逆函数。
1.1函数的概念
基本上，一个函数f有每个元素x的集合，称为Df,，有另一组的元素y，称为Rf。我们说Df是f的定义域，Rf是f的值域，用方程y = f(x)来表示他们之间的关系。习惯上称符号x为自变量，y为因变量。
-
-
- 要使Df ={a, b, c}, Rf = {1, 2, 3}且f(a) = 1, f(b) = 2，f(c) = 3
-
画出f的图像。
例1.1.2画出f(x) = |x|的图像。
设Df为所有实数集，Rf为所有非负实数集，对于Df中的每一个x，使y = |x|在集合Rf中，假若这样，f(x) = |x|，x的绝对值，回顾一下
我们记f(0) = 0, f(1) = 1 和 f(minus;1) = 1，
如果函数f的定义域Df和值域Rf都是实数集的子集，则f的图像是所有实数对 (x, f(x))，xisin;Df。这个图像可能会被画在xy平面上，使用y=f（x）。绝对值函数的图像描绘如下：
例1.1.3画出图像f(x)= 。
为了使f的范围只包含实数，我们必须假设限制x ge; 4。因此定义域Df包含所有使实数x ge; 4的集合x。值域包含所有使y ge; 0的实数y。f的图像描绘如下：
例1.1.4一个在工程中有用的函数是单位阶跃函数，u，定义如下：
u（x）的图像在x=0处有一个向上飞跃，它的图像如下：
例1.1.5画出f(x)=的图像。
很明显Df包含所有实数且xne;plusmn;2，f的图像如下：
我们注意到几件关于这个函数图像的事情。第一、这个图像有三个不同的部分，被两条虚线竖线x = minus;2 和 x = 2分隔，这些垂直的线，x = plusmn;2，被称为垂直渐近线。第二、对于x的大的正负值，f（x）趋向于0.因此，x轴和方程y=0，被称为水平渐近线。
设函数f的定义域Df和值域Rf为实数集，则对于所有的xisin;Df，若f(x) = f(minus;x)则f被称为偶函数。且对于所有的xisin;Df，若f(minus;x) = minus;f(x)，则f被称为奇函数。同样的，若x1 = x2，f(x1) = f(x2)，则f被称为一对一函数。
例1.1.6 画出函数f(x) = x4 minus; x2的图像。
这个函数是偶函数因为对于所有的x我们有
f(minus;x) = (minus;x)4 minus; (minus;x)2 = x4 minus; x2 = f(x).
函数f的图像是关于y轴对称的，因为对于每一个x，(x, f(x)) 和 (minus;x, f(x))都在图像上。偶函数的图像总是关于y轴对称。f的图像如下：
这个f函数不是一对一函数因为f(minus;1) = f(1).
例1.1.7 画出g(x) = x3 minus; 3x.的函数图像。
这个函数g是奇函数，因为对于每一个x，
g(minus;x) = (minus;x)3 minus; 3(minus;x) = minus;x3 3x = minus;(x3 minus; 3x) = minus;g(x).
函数g的图像关于原点对称，因为对于所有的x，(x, g(x)) 和 (minus;x, minus;g(x))都在图像上。一个奇函数的图像总是关于原点对称。函数g的图像如下：
这个函数不是一对一函数因为g(0)=g()=g(﹣)，可以看出每一个函数f可以被写成一个偶函数和一个奇函数。
令 g(x) = (f(x) f(minus;x)), h(x) = (f(x) minus; f(minus;x)).
则 g(minus;x) = (f(minus;x) f(x)) = g(x)
h(minus;x) = (f(minus;x) minus; f(x)) = minus;h(x).
此外 f(x) = g(x) h(x).
例1.1.8 f表示成一个偶函数和一个奇函数，当f(x) = x4 minus; 2x3 x2 minus; 5x 7.
我们定义g(x) = (f(x) f(minus;x))
{(x4 minus; 2x3 x2 minus; 5x 7) (x4 2x3 x2 5x 7)}
=x4 x2 7
和 h(x) = (f(x) minus; f(minus;x))
{(x4 minus; 2x3 x2 minus; 5x 7) minus; (x4 2x3 x2 5x 7)}
=minus;2x3 minus; 5x.
则明显地g（x）是偶函数，h（x）是奇函数。
g(minus;x) = (minus;x)4 (minus;x)2 7
=x4 x2 7
= g(x)
h(minus;x) = minus; 2(minus;x)3 minus; 5(minus;x)
=2x3 5x
=minus;h(x).
我们记g(x) h(x) = (x4 x2 7) (minus;2x3 minus; 5x)
=x4 minus; 2x3 x2 minus; 5x 7
=f(x).
讲述一个函数是否是一对一函数是不容易的。图形化测试是如果没有水平线不止一次地穿过f的图像，则f是一对一函数。数学上表示f为一对一函数，我们需要表示当x1 = x2.时，f(x1) = f(x2)。
例1.1.9 表明f(x) = x3是一对一函数
假设f(x1) = f(x2)，则
0 = x31 minus; x32
= (通过分解)
若 x1 ne;x2, 则 2= 0
和
=
如果x1不是一个实数，这是唯一可能的。这个矛盾证明了若
x1ne;x2，则f(x1) ne; f(x2)。因此，f是一对一函数。f的图像如下：
如果一个有着定义域Df和值域Rf的函数f是一对一函数，则f有一个独特的逆函数g（定义域Df和值域Rf）
这样对每一个xisin;Df ，g(f(x)) = x和对于这样的 yisin; Rf , f(g(y)) = y.
这个函数g也可以写成fminus;1 ，明确的表示g是不容易的但是接下来的算法可以帮助我们计算g：
第一步 根据y的值解方程y=f（x），求解x的值，确保存在一个解x。
第二步 写x=g(y)，当个g（y）是第一步获得的唯一解。
第三步 如果它是可取的x代表独立变量，y代表非独立变量，然后在第二步中交换x和y，写下y=g（x）。
备注一、如果y = f(x) 和y = g(x) = fminus;1(x)是画在同一坐标轴，那么，y=f（x）的图像和y=g（x）的图像是关于直线y=x的一个镜像。
例1.1.10 确定f(x) = x3.的逆函数。
我们已经从例9中知道f是一对一函数，因此，它有一个独特的逆函数，我们使用上面的算法来计算g = fminus;1.
第一步、我们求y=x3中x的值，得出x=，这是独特的解决方法。
第二步、那么 g(y) =、g(x) ==fminus;1(x).
第三步、我们在同一坐标轴上画出y=x3和y=图像，然后比较他们的图像。
一个n项的多项式函数的一般形式：
p(x) = a0xn a1xnminus;1 · · ·
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