Course: 1.16 职能的组成 | The Way To Learn

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职能的组成

If f(x) = x + 2, and g(x) = 2x + 4, what is f(g(x))?
::如果f(x) = x + 2, g(x) = 2x + 4, 什么是 f(g(x) ) ?

A function can be conceptualized as a 'black box'. The input, or x value is placed into the box, and the box performs a specific set of operations on it. Once the operations are complete, the output (the " f(x) " or " y " value) is retrieved. Once the output is retrieved, the box is ready to work on the next input.
::函数可以概念化为“黑盒 ” 。输入或 x 值被放入框中, 框将执行特定的操作。一旦操作完成, 输出( f(x) 或“y” 值) 就会被检索。一旦输出被检索, 框将准备对下一个输入工作。

Using this idea, function composition can be seen as a box inside of a box. The input x value goes into the inner box, and then the output of the inner box is used as the input of the outer box.
::使用这个概念, 函数构成可以被视为框内的一个框。输入 x 值会进入内框, 然后内框的输出会用作外框的输入。

Composition of Functions
::职能的组成

Functions are often described in terms of “input” and “output.” For example, consider the function f ( x ) = 2 x + 3. When we input an x value, we output a y value, or a function value. We find the output by taking the input x , multiplying by 2, and adding 3. We can do this for any value of x . Now consider a second function g ( x ) = 5 x . For this function too, we can take an x value, input the x into g ( x ), and obtain an output. What happens if we take the output of g and use it as the input of f ?
::函数通常用“投入”和“产出”来描述。例如,考虑函数 f(x) = 2x + 3. 当我们输入一个 x 值时,我们输出一个 y 值或一个函数值。我们通过输入 x 找到输出,乘以 2 和增加 3 。我们可以为 x 的任何值这样做。现在考虑第二个函数 g(x) = 5x 。对于这个函数,我们也可以将一个 x 值输入 g(x),将 x 输入到 g(x) 中,然后获得输出。如果我们将 g 的输出作为 f 的输入,结果会怎样?

Examples
::实例

Example 1
::例1

Earlier, you were given a problem about finding a composite function .
::早些时候,有人给了你一个寻找复合函数的问题。

If f(x) = x + 2 , and g(x) = 2x + 4 , what is f(g(x)) ?
::如果f(x) = x + 2, g(x) = 2x + 4, 什么是 f(g(x) ) ?

f(g(x)) = f(2x + 4) = (2x + 4) + 2 = 2x + 6
::f(g(xx)) = f(2x+4) = (2x+4) = (2x+4) + 2 = 2x+6

Example 2
::例2

Given the function definition above, g ( x ) = 5 x . Therefore if x = 4, then we have g (4) = 5(4) = 20. What happens if we then take the output of 20 and use it as the input of f ?
::根据上文g(x) = 5x 的功能定义,g(x) = 5x。因此,如果 x = 4,那么我们有 g(4) = 5(4) = 20。如果我们将20 的输出作为 f 的输入,结果会如何?

Substituting 20 in for x in f ( x ) = 2 x + 3 gives: f (20) = 2(20) + 3 = 43.
::在 f(x) = 2x + 3 中 x 的替代值为 20 英寸: f(20) = 2(20) + 3 = 43。

The table below shows several examples of this same process:
::下表列出了同一过程的几个例子:

x	Output from g	Output from f
2	10	23
3	15	33
4	20	43
5	25	53

Examining the values in the table, we can see a pattern : all of the final output values from f are 3 more than 10 times the initial input. We have created a new function called h ( x ) out of f ( x ) = 2 x + 3 in which g ( x ) = 5 x is the input:
::检查表格中的值,我们可以看到一个模式: f 的所有最终输出值是初始输入的10倍以上的3倍以上。我们从 f(x) = 2x + 3中创建了名为 h(x) 的新函数, 其中 g(x) = 5x 的输入为 g(x) = 5x :

h ( x ) = f (5 x ) = 2(5 x ) + 3 = 10 x + 3
::h(x) = f(5x) = 2(5x) = 2(5x) + 3 = 10x + 3

When we input one function into another, we call this the composition of the two functions. Formally, we write the composed function as f ( g ( x )) = 10 x + 3 or write it as ( f o g ) x = 10 x + 3
::当我们将一个函数输入到另一个函数时, 我们将此称为两个函数的构成。形式上, 我们将组成函数写成 f( g( x) = 10x + 3, 或者写成 (f o( g) x = 10x + 3)

Example 3
::例3

Find f ( g ( x )) and g ( f ( x )):
::查找 f( g( x)) 和 g( f( x) ) :

f (x) = 3 x + 1 and g ( x ) = x ²
::f(x) = 3x + 1 和 g(x) = x2

f ( g ( x )) = f ( x ² ) = 3( x ² ) + 1 = 3 x ² + 1
::f(g(xx)) = f(x2) = 3(x2) = 3(x2) + 1 = 3x2 + 1

g ( f ( x )) = g (3 x + 1) = (3 x + 1) ² = 9 x ² + 6 x + 1
::g(f(xx)) = g(3x+1) = (3x+1) = (3x+1) = (3x+1) 2 = 9x2 + 6x+ 1

In both cases, the resulting function is quadratic.
::在这两种情况下,由此产生的功能都是二次函数。

f (x) = 2 x + 4 and g ( x ) = (1/2) x - 2
::f(x) = 2x + 4 和 g(x) = (1/2) x - 2

f ( g ( x )) = 2((1/2) x - 2) + 4 = (2/2) x - 4 + 4 = (2/2) x = x
::f(g(xx)) = 2(2(1/2)x-2) + 4 = (2/2)x - 4 + 4 + 4 = (2/2)x = x

g ( f ( x )) = g (2 x + 4) = (1/2)(2 x + 4) - 2 = x + 2 - 2 = x .
::g(f(x)) = g(2x+4) = (1/2) (2x+4) - 2 = x+ 2-2= x。

In this case, the composites were equal to each other, and they both equal x , the original input into the function. This means that there is a special relationship between these two functions. We will examine this relationship in Chapter 3. It is important to note, however, that f ( g ( x ) is not necessarily equal to g ( f ( x )).
::在这种情况下,复合体是等同的,它们都等于函数的原始输入的x。这意味着这两种功能之间有特殊的关系。我们将在第3章中审查这种关系。但必须指出,f(g(x)不一定等于g(f(x))。

Example 4
::例4

Decompose the function f ( x ) = (3 x - 1) ² - 5 into a quadratic function g ( x ) and a linear function h ( x ).
::将函数 f(x) = (3x-1) 2 - 5 分解成四方函数 g(x) 和线性函数 h(x) 。

When we compose functions, we are combining two (or more) functions by inputting the output of one function into another. We can also decompose a function. Consider the function f ( x ) = (2 x + 1) ² . We can decompose this function into an “inside” and an “outside” function. For example, we can construct f ( x ) = (2 x + 1) ² with a linear function and a quadratic function. If g ( x ) = x ² and h ( x ) = (2 x + 1), then f ( x ) = g ( h ( x )). The linear function h ( x ) = (2 x + 1) is the inside function, and the quadratic function g ( x ) = x ² is the outside function.
::当我们组成函数时, 我们通过将一个函数的输出输入到另一个函数中, 将两个( 或更多) 函数合并。我们还可以分解一个函数。将函数 f( x) = 2x + 1 2, 我们可以将这个函数分解成“ 内侧” 和“ 外侧” 函数。例如, 我们可以构造 f( x) = 2x = 2x = 2x 2 和 h( x) = 2x + 1, 然后 f( x) = g( h) x ) 。线性函数 h( x) = 2x + 1) 是内部函数, 二次函数 g( x) = x2 是外部函数。

Let h ( x ) = 3 x - 1 and g ( x ) = x ² - 5. Then f ( x ) = g ( h ( x )) because g ( h ( x )) = g (3 x - 1) = (3 x - 1) ² - 5.
::let h(x) = 3x-1 和 g(x) = x2 - 5. 然后f(x) = g = g x) = g(3x-1) = (3x-1) 2 - 5.

The decomposition of a function is not necessarily unique. For example, there are many ways that we could express a linear function as the composition of other .
::函数的分解不一定是独一无二的。例如,我们可以以多种方式将线性函数表达为其他函数的构成。

Example 5
::例5

Given:
::参照:

$\begin{aligned} f (x) = 5 x + 3 \end{aligned}$
:xx)=5x+3

$\begin{aligned} g (x) = 3 x^{2} \end{aligned}$
::g(x)=3x2

Find: $\begin{aligned} f (g (4)) \end{aligned}$
::查找:f(g(4))

To find f ( g (4)), we need to know what $\begin{aligned} g (4) \end{aligned}$ is, so we know what to substitute into $\begin{aligned} f (x) \end{aligned}$ :
::要找到f(g(4)),我们需要知道g(4)是什么, 所以我们知道什么可以替代 f(x) :

Substitute 4 for x for the function g ( x ), giving: $\begin{aligned} 3 \cdot 4^{2} \end{aligned}$
::函数 g(x) 的 x 替代 4 , 给: 342

Simplify: $\begin{aligned} 3 \cdot 16 = 48 \end{aligned}$
::简化: 316=48

$\begin{aligned} ∴ g (4) = 48 \end{aligned}$
::*g(4)=48

Substitute 48 for the x in the function $\begin{aligned} f (x) \end{aligned}$ giving: $\begin{aligned} 5 (48) + 3 \end{aligned}$
::函数 f( x) 给付: 5( 48)+3 中 x x 的替代值为 48

Simplify: $\begin{aligned} 240 + 3 = 243 \end{aligned}$
::简化: 240+3=243

$\begin{aligned} ∴ f (g (4)) = 243 \end{aligned}$
::f(g(4))=243

Example 6
::例6

Given:
::参照:

%20%3D%207n%20%2B1%20%2B%204(g)"> $\begin{aligned} h (n) = 7 n + 1 + 4 (g (n)) \end{aligned}$
::h=7n+1+4(g)

$\begin{aligned} g (t) = - t \end{aligned}$
:gt)____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

$\begin{aligned} f (x) = - 2 x + g (x) \end{aligned}$
:xx)%%2x+g(x)

Find: $\begin{aligned} f (h (- 5)) \end{aligned}$
::查找:f(h(-5))

First, let's solve for the value of the inner function, $\begin{aligned} h (- 5) \end{aligned}$ . Then we'll know what to plug into the outer function.
::首先,让我们解决内部函数, h(- 5) 的值, 然后我们就会知道在外函数中插入什么。

$\begin{aligned} h (- 5) = (7) (- 5) + 1 + 4 (g (- 5)) \end{aligned}$
::h(-5)=(7)(-5)+1+4(g(-5))

To solve for the value of h , we need to solve $\begin{aligned} g (- 5) \end{aligned}$
::要解决 h 值, 我们需要解决 g( - 5) 。

$\begin{aligned} g (- 5) = - (- 5) \end{aligned}$
::g(-5) (-5)

$\begin{aligned} ∴ g (- 5) = 5 \end{aligned}$
::*g(-5)=5

Now we have: $\begin{aligned} h (- 5) = (7) (- 5) + 1 + (4) (5) \end{aligned}$
::现在我们有:h(-5)=(7)(-5)+1+(4)(5)

Simplify to get: $\begin{aligned} h (- 5) = - 14 \end{aligned}$
::简化以获得: h(- 5) @% 14

Now we know that $\begin{aligned} h (- 5) = - 14 \end{aligned}$ . That tells us that $\begin{aligned} f (h (- 5)) \end{aligned}$ is $\begin{aligned} f (- 14) \end{aligned}$
::也就是说f(h(-5))是f(-14)

Find $\begin{aligned} f (- 14) = (- 2) (- 14) + g (- 14) \end{aligned}$
::查找 f( - 14) = (-2)( - 14)+g( - 14)

So to solve for the value of $\begin{aligned} f (- 14) \end{aligned}$ , we need to solve for the value of $\begin{aligned} g (- 14) \end{aligned}$
::因此,要解决f(-14)的值,我们需要解决g(-14)的值

$\begin{aligned} g (- 14) = - (- 14) \end{aligned}$
::g(-14) (-14)

$\begin{aligned} ∴ g (- 14) = 14 \end{aligned}$
::*g(-14)=14

Now we can finish up!
::现在我们可以完成!

$\begin{aligned} f (- 14) = (- 2) (- 14) + 14 \end{aligned}$
::f(-14)=(-2)(-14)+14

$\begin{aligned} ∴ f (- 14) = 42 \end{aligned}$
::f( - 14) = 42

Review
::回顾

For problems 1-4:
::对于问题1-4:

$\begin{aligned} f (x) = 2 x - 1 \end{aligned}$ $\begin{aligned} g (x) = 3 x \end{aligned}$ $\begin{aligned} h (x) = x^{2} + 1 \end{aligned}$
:x)=2x-1g(x)=3xh(x)=x2+1

Find: $\begin{aligned} f (g (- 3)) \end{aligned}$
::查找: f( g( - 3) )

Find: $\begin{aligned} f (h (7)) \end{aligned}$
::查找:f(h(7))

Find: $\begin{aligned} h (g (- 4)) \end{aligned}$
::查找:h(g(- 4))

Find: $\begin{aligned} f (g (h (2))) \end{aligned}$
::查找:f(g(h(2)))

Evaluate each composition below:
::评估以下每一组成:

Given: $\begin{aligned} f (x) = - 5 x + 2 \end{aligned}$ and $\begin{aligned} g (x) = \frac{1}{2} x + 4 \end{aligned}$ . Find $\begin{aligned} f (g (12)) \end{aligned}$ .
::给定 : f( x) =% 5x+2 和 g( x) = 12x+4。查找 f( g( 12) ) 。

Given: $\begin{aligned} g (x) = - 3 x + 6 \end{aligned}$ and $\begin{aligned} h (x) = 9 x + 3 \end{aligned}$ . Find $\begin{aligned} g (h (\frac{1}{3})) \end{aligned}$ .
::给定 : g( x) % 3x+6 和 h( x) = 9x+3. 查找 g( h) (13) 。

Given: $\begin{aligned} f (x) = - \frac{1}{5} x + 4 \end{aligned}$ and $\begin{aligned} g (x) = 4 x^{2} \end{aligned}$ . Find $\begin{aligned} f (g (10)) \end{aligned}$ .
::给定 : f( x) 15x+4 和 g( x) = 4x2. 查找 f( g(10) ) 。

Given $\begin{aligned} g (x) = 3 | x - 4 | + 6 \end{aligned}$ and $\begin{aligned} h (x) = - x^{3} \end{aligned}$ . Find $\begin{aligned} h (g (4)) \end{aligned}$ .
::g( x) = 3x-46 和 h( x) x3. 查找 h( g(4) ) 。

Given $\begin{aligned} f (x) = \sqrt{x + 2} \end{aligned}$ and $\begin{aligned} g (x) = | 2 x | \end{aligned}$ . Find $\begin{aligned} g (f (- 7)) \end{aligned}$ .
::给定 f( x) =x+2 和 g( x)\\\\\\\\\\\\\\\\\ 查找 g( f( f) - 7) 。

Given $\begin{aligned} f (x) = - 3 x + 2 \end{aligned}$ and given $\begin{aligned} g (x) = 2 x^{2} \end{aligned}$ and given $\begin{aligned} h (x) = 4 | 7 - x | + 6 \end{aligned}$ . Find $\begin{aligned} f (g (h (1))) \end{aligned}$ .
::给定 f( x) =% 3x+2, 给定 g( x) = 2x2, 给定 h( x) = 4 @ @ 7- x# 查找 f( g( h(1) ) ) 。

Given $\begin{aligned} f (x) = (- 3) \end{aligned}$ and given $\begin{aligned} g (x) = \sqrt{2 x} \end{aligned}$ and given $\begin{aligned} h (x) = | 4 x | - 12 \end{aligned}$ . Find $\begin{aligned} f (h (g (18))) \end{aligned}$ .
::给定 f( x) = (-3) , 给定 g( x) = 2x , 给定 h( x) = 4x = 12。查找 f( h) ( g( 18) ) 。

Are compositions commutative? In other words, does $\begin{aligned} f (g (x)) = g (f (x)) \end{aligned}$ ?
::构成是否具有通融性?换句话说,f(g(x))=g(f(x))?

Given: $\begin{aligned} f (x) = - 2^{2} - 5 x \end{aligned}$ and $\begin{aligned} h (x) = 3 x + 2 \end{aligned}$ . Find $\begin{aligned} f (h (x)) \end{aligned}$ .
::给定: f( x)\\% 22- 5x 和 h( x) = 3x+2. 查找 f( h( x) ) 。

Two functions are inverses of each other if $\begin{aligned} f (g (x)) = x \end{aligned}$ and $\begin{aligned} g (f (x)) = x \end{aligned}$ If $\begin{aligned} f (x) = x + 3 \end{aligned}$ , find its inverse: $\begin{aligned} g (x) \end{aligned}$
::如果 f( g( x)) =x 和 g( f( f( x) ) =x 如果 f( x) =x+3 发现其反义 : g( x)

A toy manufacturer has a new product to sell. The number of units to be sold, n , is a function of the price p such that: $\begin{aligned} n (p) = 30 - 25 p \end{aligned}$ . The revenue r earned from the sales is a function of the number of units sold n such that: %20%3D%201000%20-%20%5Cfrac%7B1%7D%7B4%7Dn%5E2"> $\begin{aligned} r (n) = 1000 - \frac{1}{4} n^{2} \end{aligned}$ . Find the function for revenue in terms of price, p .
::玩具制造商有一个新的产品要出售。要出售的单位数目 n 取决于价格p, 即:n(p)=30-25p。销售所得的收入是售出的单位数目的函数,即:r=1000-1400-1400n2。从价格上确定收入的函数,p。

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

职能的组成

Composition of Functions ::职能的组成

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Review ::回顾

Review (Answers) ::回顾(答复)