简单复合函数

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The Purpose of This Lesson
::本课程的目的

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In this lesson, you will explore combinations of functions called composite functions.
::在这一教训中,你们将探讨综合职能的组合。

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Introduction: Composing an Absolute Value Function and a Linear Function
::导言:包含绝对值函数和线性函数

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Consider the diagram below. The diagram represents a top view of the scenario. The units on the axes are feet. Rafael is standing at point A with a flashlight. A mirror is placed along the $\begin{array}{r}x\end{array}$ -axis at point B. Rays of light travel in straight lines. Find the equation for the line representing the path of light from Rafael to the mirror. The line graphed shows the path the light would take if it passed through the mirror.  An a bsolute value function can be used to model the path of reflected light. The path is shown in the graph below.
::考虑下面的图表。 图表代表了情景的顶部视图。 轴上的单位是脚。 Rafael 站在 A 点, 手电筒在A 点。 在 B 点, 光线的光线以直线沿 X 轴线放置一个镜像。 查找代表光线路径的线的方程式, 从 Rafael 到镜子。 以图形显示的线条显示光线通过镜片时将走的路。 绝对值函数可用于模拟反光路径。 路径在下面的图表中显示 。

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The light takes a path from point A to a mirror at point B. This is a linear path up until point B . The orange line shows the path the light would take if it continued. The equation for  the orange line is: 
::光从 A 点到 B 点的镜像路径。 这是一直到 B 点的线性路径。 橙色线条显示光线如果继续的话会走的路径。 橙色线的方程式是 :

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::y* 59x+10

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C hanging this function to an absolute value function changes the negative  $\begin{array}{r}y\end{array}$ -values and makes them positive, modeling the reflection of light by  a mirror. The equation for the absolute value function is:
::将此函数更改为绝对值函数, 改变负 Y 值, 使其为正值, 用镜子模拟光的反射。 绝对值函数的方程式是 :

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::y"59x+10"\\\\\\"59x+10\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"59x+10\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"59x+10\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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Activity 1 : Composites of Absolute Value Functions and Other Functions
::活动1:绝对值函数和其他函数的组合

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Interactive
::交互式互动

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Use the interactive below to experiment with changing the parameters for $\begin{array}{r}f(x).\end{array}$  Predict the graph of  $\begin{array}{r}\left|f(x)\right|,\end{array}$ then confirm your results by clicking the checkbox to view the composite function. Are there any versions of  $\begin{array}{r}f(x)\end{array}$  for which the composite is the same as the original? Explain.  
::使用下面的交互效果来试验f(x) 的参数。 预测 \\\ f(x)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

 

 

 

 

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Interactive
::交互式互动

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Use the interactive to experiment with the parameters for $\begin{array}{r}f(x).\end{array}$  Predict the effect of the composition of the absolute value function and  $\begin{array}{r}\left|f(x)\right|.\end{array}$  Click to view  $\begin{array}{r}\left|f(x)\right|.\end{array}$  
::使用交互性来实验 f( x) 的参数。 预测绝对值函数和\\ f( x)\\\\\\\\\\\\\\\\\\\\\ x\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

 

 

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Work it Out
::工作出来

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Given the graphs of each  $\begin{array}{r}f(x)\end{array}$  below, visualize and sketch the graph of  $\begin{array}{r}\left|f(x)\right|.\end{array}$
::根据下面每个f(x)的图表,可视化和草图显示 f(x)的图表。

1.  

   

2.  

   

3.  

   

    
4.  

   

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Activity 2:  Evaluating Composite Functions
::活动2:评价综合职能

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The composition of two functions means that the output from one function is substituted as the input for another function. 
::两个职能的组成意味着将一个职能的产出替换为另一个职能的输入。

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For example, if $\begin{array}{r}f(x)=x+3\end{array}$  and $\begin{array}{r}g(x)=2x,\end{array}$   we can  find the composition of $\begin{array}{r}g(f(3))\end{array}$  like so:
::例如,如果f(x)=x+3和g(x)=2x,我们能找到g(f(3))的成分,例如:

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::f( x) =x+3 某些输入 x 的函数 f 向输入增加 3 。 f(3) = (3)+3 3 输入 3, x. f(3)=6 输入, f的输出为 6. g( x) = 2xx 输入 。 某些输入 x 的函数 g 乘以 2. g( f(3)) =2( f(3)) 的输出为 g( x) g. g( f(3)) =2(6) 输入 6, f(3). g( f(3)) =12 输入时, g的输出为 12。 f(3) 输入时, g的输出为 12 。

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Work it Out
::工作出来

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1. The last oval in the  function map   below  represents   $\begin{array}{r}g\left(f(x)\right).\end{array}$   You should read this " $\begin{array}{r}g\end{array}$  of  $\begin{array}{r}f\end{array}$  of  $\begin{array}{r}x\end{array}$ ." Use the function map to  evaluate  the following:  
   ::函数映射下面的最后一个 oval 代表 g(f(x) ) 。 您应该读取“ x 的 f 的 g ” 。 使用该函数映射来评估以下内容 :

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$\begin{array}{r}\begin{array}{ll}\text{a.}& f(2)\\ \text{b.}& g(f(2))\\ \text{c.}& g(6)\\ \text{d.}& g(7)\\ \text{e.}& g(f(3))\\ \text{f.}& f(4)\\ \text{g.}& g(f(5))\end{array}\end{array}$
::af(2)b.g(f(2))c.g(6)d.g(7)例如(f(3))f.f(4)g(f(5))

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2. In part g) of the last problem,  you were asked to find  $\begin{array}{r}g(f(5)).\end{array}$  This wasn't possible because 5 isn't in the domain for  $\begin{array}{r}f.\end{array}$  The output of  $\begin{array}{r}f(5)\end{array}$  doesn't exist, because the domain, $\begin{array}{r}x,\end{array}$  only includes the values 1 through 4. That means  $\begin{array}{r}g(f(5))\end{array}$  doesn't exist. What is the domain of  $\begin{array}{r}g(f(x))?\end{array}$  Explain.
   ::在最后一个问题的 g 部分中, 您被要求查找 g( f (5)) 。 这是不可能的, 因为 5 不属于 f 的域 。 f (5) 的输出不存在, 因为 f (5) 的输出只包含 1 到 4 的值, 这意味着 g( f (5)) 不存在。 g ( f ( x) 的域是什么? 解释 。
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3. (extension) Given any composition of functions  $\begin{array}{r}g(f(x)),\end{array}$  the domain is restricted to the domain of  $\begin{array}{r}f(x),\end{array}$  and also restricted to the values of  $\begin{array}{r}x\end{array}$  such that  $\begin{array}{r}f(x)\end{array}$  is in the domain of  $\begin{array}{r}g.\end{array}$  Given $\begin{array}{r}g(x)=\sqrt{x}\end{array}$  and $\begin{array}{r}f(x)=\frac{1}{x},\end{array}$  use the composition  $\begin{array}{r}g(f(x))\end{array}$  to explain the previous statement.
   :extension) 鉴于函数 g(f(x) 的任何构成, 域限 f(x) 的域, 并且也限制于 x 的值, 使 f(x) 位于 g. g. g(x) =x 和 f(x) = 1x 的域, 使用 compress g(f(x) ) 来解释前一个语句 。

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Activity 3: Building Composite Functions
::活动3:建立综合职能

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Remember in , $\begin{array}{r}x\end{array}$  is the input and $\begin{array}{r}f(x)\end{array}$  is the output. In order to find a composite function , take the output of a function and substitute it as the input  of another function. 
::记住, x 是输入, f(x) 是输出。为了找到一个复合函数,请使用函数的输出,将其替换为另一个函数的输入。

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Consider  the functions   $\begin{array}{r}f(x)=x+2\end{array}$  and $\begin{array}{r}g(x)=2x+4,\end{array}$  for example. The composition  $\begin{array}{r}f\end{array}$  of  $\begin{array}{r}g\end{array}$  of  $\begin{array}{r}x,\end{array}$   or $\begin{array}{r}f(g(x))\end{array}$  means that the output of $\begin{array}{r}g(x)\end{array}$   will be  the input  of   $\begin{array}{r}f(x).\end{array}$  In other words,  substitute   $\begin{array}{r}g(x)\end{array}$  for $\begin{array}{r}x\end{array}$  in $\begin{array}{r}f(x),\end{array}$  as shown below: 

::函数 f( x) =x+2 和 g( x) = 2x+4 。 例如, 函数 f( x) 和 g( x) = 2x+4 。 g( x) 的构成f x, orf( g( x) 的构成f f) 表示 g( x) 的输出将是 f( x) 的输入。 换句话说, f( x) 中的 x 替代 g( x) 。 如下: f( g( x) = f( 2x+4) = ( 2x4) +2= 2x+6

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Example 3 -1
::例3-1

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Let   $\begin{array}{r}f(x)=5x\end{array}$  and $\begin{array}{r}g(x)=2x+3.\end{array}$   Eval ua te   $\begin{array}{r}g(f(4)).\end{array}$   Find $\begin{array}{r}g(f(x))\end{array}$  and plug 4 into the resulting equation to confirm your answer  for   $\begin{array}{r}g(f(4)).\end{array}$  
::let f( x) = 5x 和 g( x) = 2x+3. 评价 g( f(4)) 。 查找 g( f( x)) 并在结果方程式中插入 4 以确认 g( f(4)) 的答案 。

 

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Solution: When evaluating composite functions, start from the inside and work your way out. Since  $\begin{array}{r}f\end{array}$  is the inside function, we start by plugging 4 into $\begin{array}{r}f(x),\end{array}$  which is $\begin{array}{r}f(4)=5(4)=20.\end{array}$   Then substitute 20 for $\begin{array}{r}x\end{array}$  in $\begin{array}{r}g(x),\end{array}$  which is  $\begin{array}{r}g(20)=2(20)+3=43.\end{array}$  
::解答 : 在评价复合函数时, 从内部开始, 并运行您的出路 。 由于 f 是内部函数 。 由于 f 是内部函数, 我们从将 4 插入 f( x) 开始, 即 f(4)= 5(4)=20。 然后将 x 替换为 g( x) 中的 20 , 即 g( 20) =2( 20)+3=43 。

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To find the  composite function $\begin{array}{r}g(f(x)),\end{array}$  substitute $\begin{array}{r}f(x)\end{array}$  for $\begin{array}{r}x\end{array}$  in  $\begin{array}{r}g(x).\end{array}$  
::要找到复合函数 g(f(x) ) , 用 f(x) 替换 g(x) x 。

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::g(f(x))=g(5x)=2(5x)+3=10x+3。

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Using the composite function, we can see that $\begin{array}{r}g(f(4))=10(4)+3=43,\end{array}$  which confirms our answer.
::使用复合函数,我们可以看到 g(f(4))=10(4)+3=43,这证实了我们的回答。

 

 

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Work it Out
::工作出来

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1. Given $\begin{array}{r}f(x)=2x-1,g(x)=3x,\text{ and }h(x)=x^2+1,\end{array}$  find:
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   1. $\begin{array}{r}f(g(-3))\end{array}$  
      :g( 3) )
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   2. $\begin{array}{r}f(h(7))\end{array}$  
      ::f(h(7))
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   3. $\begin{array}{r}h(g(-4))\end{array}$  
      :g(- 4))
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   4. $\begin{array}{r}f(g(h(2)))\end{array}$  
      :fg(h(2)))
   
   ::给定 f( x) = 2x-1, g( x) = 3x, h( x) =x2+1, 查找: f( g( 3) ) f( h(7)) h( g( 4)) f( g( h(2)) )
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2. Given $\begin{array}{r}f(x)=-5x+2\text{ and }g(x)=\frac{1}{2}x+4,\end{array}$  find  $\begin{array}{r}f(g(x)).\end{array}$  
   ::给定 f( x) =% 5x+2 和 g( x) = 12x+4, 请查找 f( g( x) ) 。
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3. Given $\begin{array}{r}g(x)=-3x+6\text{ and }h(x)=9x+3,\end{array}$  find  $\begin{array}{r}g(h(x)).\end{array}$  
   ::g( x) =% 3x+6 和 h( x)= 9x+3, 查找 g( h( x) ) 。
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4. Given $\begin{array}{r}f(x)=-\frac{1}{5}x+4\text{ and }g(x)=4x^2,\end{array}$  find  $\begin{array}{r}g(f(x)).\end{array}$  
   ::给定 f( x) 15x+4 和 g( x) = 4x2, 查找 g( f( x) ) 。
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5. Given $\begin{array}{r}f(x)=-3x+2\text{ and }g(x)=2x^2,\end{array}$  find  $\begin{array}{r}f(g(x)).\end{array}$  
   ::给定 f( x) 3x+2 和 g( x) = 2x2, 找到 f( g( x) ) 。
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6. A toy manufacturer has a new product to sell. The number of units to be sold, $\begin{array}{r}n,\end{array}$  is a function of the price $\begin{array}{r}p\end{array}$  such that: $\begin{array}{r}n(p)=100-5p.\end{array}$  The revenue $\begin{array}{r}r\end{array}$  is price times number of units sold, or  $\begin{array}{r}r=p\cdot n.\end{array}$  
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   1. Find the function for revenue in terms of price,  $\begin{array}{r}p.\end{array}$  
      ::以物价来寻找收入的函数,p.。
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   2. Use the function you found in part a) to calculate  the revenue if the price is $8.
      ::使用您在 a 部分找到的函数计算价格为 8 美元的收入 。
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   3. Use the function you found in part a) to calculate the revenue if the price is $11.
      ::如果价格为 11 美元, 请使用在 a 部分中找到的函数计算收入 。
   
   ::玩具制造商有一个新的产品要出售。 要出售的单位数量 n, 是价格 p 的函数, 包括: n( p) = 100-5p。 收入 r 是售出单位的价格乘数, 或 r= pn。 找到以价格计的收入的函数, p. 使用在a部分中找到的函数) 来计算价格为 8 美元的收入。 如果价格为 11 美元, 请使用在 a 部分中找到的函数来计算收入 。