节: 职能的组成 | 1.9 职能的组成

章节大纲

If $\begin{aligned} f (x) = x + 2 \end{aligned}$ , and $\begin{aligned} g (x) = 2 x + 4 \end{aligned}$ , what is $\begin{aligned} f (g (x)) \end{aligned}$ ?
::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 $\begin{aligned} x \end{aligned}$ value is placed into the box, and the box performs a specific set of operations on $\begin{aligned} x \end{aligned}$ . Once the operations are complete, the output (the “ $\begin{aligned} f (x) \end{aligned}$ ” or “ $\begin{aligned} y \end{aligned}$ ” value) is provided. Once the output is provided, the box is ready to work on the next input.
::函数可以被概念化为`黑盒 ' 。输入或 x 值被放入框中,框在 x 上执行一套特定的操作。操作完成后,即提供输出(“f(x)”或“y”值)。一旦输出提供,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 $\begin{aligned} x \end{aligned}$ 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。

Function Composition
::职能构成构成

Functions are often described in terms of “input” and “output”. For example, consider the function $\begin{aligned} f (x) = 2 x + 3 \end{aligned}$ . When we input an $\begin{aligned} x \end{aligned}$ value, we output a $\begin{aligned} y \end{aligned}$ value, or a function value. We find the output by taking the input $\begin{aligned} x \end{aligned}$ , multiplying by 2, and adding 3. We can do this for any value of $\begin{aligned} x \end{aligned}$ . Now consider a second function $\begin{aligned} g (x) = 5 x \end{aligned}$ . For this function too, we can take an $\begin{aligned} x \end{aligned}$ value, input the $\begin{aligned} x \end{aligned}$ into $\begin{aligned} g (x) \end{aligned}$ , and obtain an output. What happens if we take the output of $\begin{aligned} g \end{aligned}$ and use it as the input of $\begin{aligned} f \end{aligned}$ ?
::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 值或函数值。我们通过输入 x 来发现输出,乘以2 和增加 3 。我们可以为任何 x 值这样做。现在考虑第二个函数 g(x)=5x。对于这个函数,我们可以将一个 x 值输入 g(x),然后获得输出。如果我们将 g 的输出作为 f 的输入,会发生什么?

Given the function definition above, $\begin{aligned} g (x) = 5 x \end{aligned}$ . Therefore if $\begin{aligned} x = 4 \end{aligned}$ , then we have $\begin{aligned} g (4) = 5 (4) = 20 \end{aligned}$ . What happens if we then take the output of 20 and use it as the input of $\begin{aligned} f \end{aligned}$ ?
::根据上文 g(x) = 5x 的函数定义, g(x) = 5x。因此, 如果 x= 4, 那么我们就有 g(4) = 5(4) = 20。如果我们以 20 的输出作为 f 的输入, 那么结果会如何 ?

Substituting 20 in for $\begin{aligned} x \end{aligned}$ in $\begin{aligned} f (x) = 2 x + 3 \end{aligned}$ gives: $\begin{aligned} f (20) = 2 (20) + 3 = 43 \end{aligned}$ .
::在 f(x) = 2x+3 中, x 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:

$\begin{aligned} x \end{aligned}$ ::x x	Output from $\begin{aligned} g \end{aligned}$ ::g 输出输出	Output from $\begin{aligned} f \end{aligned}$ ::f 产出产出 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 $\begin{aligned} f \end{aligned}$ are 3 more than 10 times the initial input. We have created a new function called $\begin{aligned} h (x) \end{aligned}$ out of $\begin{aligned} f (x) = 2 x + 3 \end{aligned}$ in which $\begin{aligned} g (x) = 5 x \end{aligned}$ is the input:
::检查表格中的值, 我们可以看到一个模式: f 的全部最终输出值是初始输入的10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 是输入 :

\begin{aligned} h (x) = f (5 x) = 2 (5 x) + 3 = 10 x + 3 \end{aligned}

::h(x)=f(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 $\begin{aligned} f (g (x)) = 10 x + 3 \end{aligned}$ or write it as $\begin{aligned} (f \circ g) x = 10 x + 3 \end{aligned}$ .
::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。正式地, 我们将组成函数写成 f( g( x)) = 10x+3 , 或者把它写成 f( fg) x= 10x+3 。

Take the two functions: $\begin{aligned} f (x) = 2 x + 4 \end{aligned}$ and $\begin{aligned} g (x) = (\frac{1}{2}) x - 2 \end{aligned}$ . Find $\begin{aligned} f (g (x)) \end{aligned}$ and $\begin{aligned} g (f (x)) \end{aligned}$ .
::采取两个函数 : f( x) = 2x+4andg( x) = ( 12) x-2 查找 f( g( x) ) 和 g( f( x) ) 。

$\begin{aligned} f (x) = 2 x + 4 \end{aligned}$ and $\begin{aligned} g (x) = (\frac{1}{2}) x - 2 \end{aligned}$
:xx) = 2x+4 和 g(x) = (12) x-2

\begin{aligned} f (g (x)) & = 2 ((\frac{1}{2}) x - 2) + 4 = (\frac{2}{2}) x - 4 + 4 = (\frac{2}{2}) x = x \\ g (f (x)) & = g (2 x + 4) = (\frac{1}{2}) (2 x + 4) - 2 = x + 2 - 2 = x . \end{aligned}

::f(g(xx))=2((12x-2)+4=(22x)x-4+4=(22x)xxx(f(x)x))=g(2x+4)=(12)(2x+4)-2=x+2-2=x。

In this case, the composites are equal to each other, and they both equal $\begin{aligned} x \end{aligned}$ , the original input into the function. This means that there is a special relationship between these two functions. We will examine this relationship in later concepts. It is important to note, however, that $\begin{aligned} f (g (x)) \end{aligned}$ is not necessarily equal to $\begin{aligned} g (f (x)) \end{aligned}$ .
::在这种情况下,复合物彼此平等,它们都等于函数的原始输入x。这意味着这两个函数之间存在特殊关系。我们将在以后的概念中研究这种关系。但必须指出,f(g(x)不一定等于g(f(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 $\begin{aligned} f (x) = (2 x + 1) 2 \end{aligned}$ . We can decompose this function into an “inside” and an “outside” function. For example, we can construct $\begin{aligned} f (x) = (2 x + 1) 2 \end{aligned}$ with a linear function and a quadratic function . If $\begin{aligned} g (x) = x 2 \end{aligned}$ and $\begin{aligned} h (x) = (2 x + 1) \end{aligned}$ , then $\begin{aligned} f (x) = g (h (x)) \end{aligned}$ . The linear function $\begin{aligned} h (x) = (2 x + 1) \end{aligned}$ is the inside function, and the quadratic function $\begin{aligned} g (x) = x 2 \end{aligned}$ 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 。我们可以将此函数分解成“ 内侧” 和“ 外侧” 函数。例如, 我们可以构造 f( x) = (2x+1) 2, 带有线性函数和四方函数。如果 g( x) =x2 和 h( x) = (2x+1) , 那么 f( x) = g( h) x) 。线性函数 h( x) = (2x+1) 是内部函数, 而四方函数 g( x) =x2 是外部函数。

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。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find $\begin{aligned} f (g (x)) \end{aligned}$ if $\begin{aligned} f (x) = x + 2 \end{aligned}$ , and $\begin{aligned} g (x) = 2 x + 4 \end{aligned}$ .
::早些时候,如果f(x)=x+2和g(x)=2x+4,您被要求找到 f(g(x)x) 。

$\begin{aligned} f (g (x)) = f (2 x + 4) = (2 x + 4) + 2 = 2 x + 6 \end{aligned}$
::f( g( x)) = f( 2x+4) = ( 2x+4) +2= 2x+6

Once you get the idea, composite functions aren’t as difficult as they look!
::综合功能不会像看起来那么困难!

Example 2
::例2

Given: $\begin{aligned} f (x) = 5 x + 3 \end{aligned}$ and $\begin{aligned} g (x) = 3 x^{2} \end{aligned}$
::给定值: f( x) =5x+3 和 g( x) = 3x2

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

To find $\begin{aligned} f (g (4)) \end{aligned}$ , 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 $\begin{aligned} x \end{aligned}$ for the function $\begin{aligned} g (x) \end{aligned}$ , 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 $\begin{aligned} x \end{aligned}$ 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 3
::例3

Given : %3D7n%2B1%2B4(g)"> $\begin{aligned} h (n) = 7 n + 1 + 4 (g (n)) \end{aligned}$ , $\begin{aligned} g (t) = - t \end{aligned}$ , and $\begin{aligned} f (x) = - 2 x + g (x) \end{aligned}$
::给定 : h( n) = 7n+1+4( g( n) ) g( t) t 和 f( x) 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) 的价值问题, 然后我们就会知道外部函数中要插入什么。

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

To solve for the value of $\begin{aligned} h \end{aligned}$ , 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) \\ ∴ g (- 5) & = 5 \end{aligned}

::g(-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) \\ ∴ g (- 14) & = 14 \end{aligned}

::g( - 14) ( - 14) ( - 14) g( - 14) =14

Now we can finish up!
::现在我们可以完成!

\begin{aligned} f (- 14) & = (- 2) (- 14) + 14 \\ ∴ f (- 14) & = 42 \end{aligned}

::f( - 14) = (-2) (-14) +14f(-14) = 42

Example 4
::例4

Given: $\begin{aligned} g (x) = 5 x^{2} \end{aligned}$ and $\begin{aligned} h (x) = 5 x^{2} - 2 x - 4 (g (x)) \end{aligned}$
::g(x)=5x2和h(x)=5x2-2-2x-4(g(x))

Find: $\begin{aligned} h (g (- 1)) \end{aligned}$
::查找: h( g( - 1))

First, solve for the value of the inner function $\begin{aligned} g (- 1) \end{aligned}$ to find what to plug into the outer function $\begin{aligned} h (g (- 1)) \end{aligned}$
::首先,解决内部函数 g(-1) 的值, 以找到在外部函数 h( g( 1) ) 中插入什么

\begin{aligned} g (- 1) & = 5 (- 1)^{2} \\ g (- 1) & = 5 \cdot 1 \\ ∴ g (- 1) & = 5 \end{aligned}

::g( - 1) =5( - 1) 2g( - 1) = 5- 1) \\\ \\ \ g( 1) =5

Next, solve for $\begin{aligned} h (g (- 1)) \end{aligned}$ which we now know is: $\begin{aligned} h (5) \end{aligned}$
::下一步, h( g( - 1)) 的解决方案, 我们现在知道是: h(5) 。

$\begin{aligned} h (5) = 5 (5^{2}) + (- 2) (5) - 4 (g (5)) \end{aligned}$
::h(5)=5(52)+(-2)(5)-4(g(5))

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

\begin{aligned} g (5) & = 5 (5^{2}) \\ g (5) & = 125 \\ ∴ h (5) & = 5 (5^{2}) + (- 2) (5) + (- 4) (125) \end{aligned}

::g(5)=5(52)g(5)=125h(5)=5(52)+(-2)(5)+(-4)(125)

Finally: $\begin{aligned} h (5) = - 385 \end{aligned}$
::最后:h(5)____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Review
::回顾

For #1-4, given $\begin{aligned} f (x) = 2 x - 1 \end{aligned}$ and $\begin{aligned} g (x) = 3 x \end{aligned}$ and $\begin{aligned} h (x) = x^{2} + 1 \end{aligned}$ , find the indicated function.
::对于 # 1-4, 给定 f( x) = 2x- 1 和 g( x) = 3x 和 h( x) = x2+ 1, 找到指定的函数。

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

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

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

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

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) = 3 x- 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) =% 2x = 查找 g( 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 和给定 g( x)= 2x2 和给定 h( x)= 4\\\\\\\\ x#6 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) 和给定 g(x) = 2x 和给定 h(x) = 4x = 12 Find f 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 $\begin{aligned} f (h (x)) \end{aligned}$
::给出的 f( x)\\\\\\\\\\\\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) , 则两个函数是反义。

A toy manufacturer has a new product to sell. The number of units to be sold, $\begin{aligned} n \end{aligned}$ , is a function of the price $\begin{aligned} p \end{aligned}$ such that: $\begin{aligned} n (p) = 30 - 25 p \end{aligned}$ . The revenue $\begin{aligned} r \end{aligned}$ earned from the sales is a function of the number of units sold $\begin{aligned} n \end{aligned}$ such that: %3D1000%20-%20%5Cfrac%7B1%7D%7B4%7Dx%5E2"> $\begin{aligned} r (n) = 1000 - \frac{1}{4} x^{2} \end{aligned}$ Find the function for revenue in terms of price, $\begin{aligned} p \end{aligned}$ .
::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=1000-14x2,从价格上找出收入的函数,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。

章节大纲

Function Composition ::职能构成构成

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)