课程： 6.9 职能的组成

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职能的组成
You are growing bacteria in a lab. Their growth varies depending on the temperature in the lab. You program the thermostat in the lab to change the temperature in the lab depending on the time of day. Since the growth of the bacteria depends on the temperature and the temperature depends on the time, is there a way to express how the growth of the bacteria depends on the time? There is. It is called composition and we consider it in this section.
::您正在实验室中生长细菌。细菌的生长因实验室的温度而异。您在实验室中制作了自动调温器, 以便根据时间改变实验室的温度。由于细菌的生长取决于温度, 温度取决于时间, 是否有一种方法来表达细菌的生长取决于时间? 有。它被称为构成, 我们在本节中考虑它。

Composition of Functions
::职能的组成

Unlike the operations we considered in the previous section that we can also perform on numbers, composition is an operation that can only be performed on functions. Composition is the operation where we input one function into another. We call this composing two functions.
::与我们在前一节中考虑的我们也可以用数字执行的行动不同,组成是只能凭职能才能执行的行动。组成是将一个功能输入到另一个功能的行动。我们称之为由两个功能组成的行动。

The symbol for composition is an open circle, $\begin{aligned} \circ \end{aligned}$ . To indicate composition, we write $\begin{aligned} (f \circ g) (x) \end{aligned}$ , which is equal to $\begin{aligned} f (g (x)) \end{aligned}$ . Notice, since $\begin{aligned} g (x) \end{aligned}$ is inside the " data-term="Parentheses" role="term" tabindex="0"> parentheses for f , it is the input. We read $\begin{aligned} (f \circ g) (x) \end{aligned}$ as “ $\begin{aligned} f \end{aligned}$ of $\begin{aligned} g \end{aligned}$ of $\begin{aligned} x \end{aligned}$ ."
::构成的符号是一个开放的圆,\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Just like subtraction and division , the order that you write the composition in is important and cannot be switched. While $\begin{aligned} (f \circ g) (x) = (g \circ f) (x) \end{aligned}$ is possible, it is not guaranteed.
::和减法和除法一样,写成组成内容的顺序很重要,不能被切换。(fg)(x)=(gf(x))是可能的,但无法保证。

Let’s do an example.
::让我们举个例子。

Example 1
::例1

Using $\begin{aligned} f (x) = \sqrt{x - 8} \end{aligned}$ and $\begin{aligned} g (x) = \frac{1}{2} x^{2} \end{aligned}$ , find $\begin{aligned} (f \circ g) (x) \end{aligned}$ and $\begin{aligned} (g \circ f) (x) \end{aligned}$ .
::使用 f( x) =x-8 和 g( x) = 12x2, 查找 (fg) (x) 和 (gf)(x) 。

Solution: For $\begin{aligned} (f \circ g) (x) \end{aligned}$ , we are going to put $\begin{aligned} g (x) \end{aligned}$ into $\begin{aligned} f (x) \end{aligned}$ everywhere there is an $\begin{aligned} x \end{aligned}$ -value.
::解答 : 对于 (fg)(x), 我们要将 g(x) 输入到 f(x) 中, 任何地方都有 x 值。

$\begin{aligned} (f \circ g) (x) = f (g (x)) = f (\frac{1}{2} x^{2}) = \sqrt{\frac{1}{2} x^{2} - 8} \end{aligned}$

:f*g(x)=f(g(x))=f(12x2)=12x2=12x2-8)

Now, to find $\begin{aligned} (g \circ f) (x) \end{aligned}$ , we would put $\begin{aligned} f (x) \end{aligned}$ into $\begin{aligned} g (x) \end{aligned}$ everywhere there is an $\begin{aligned} x \end{aligned}$ -value.
::现在,为了找到 (gf(x)),我们将将 f(x) 放到 g(x) 中, 任何地方都有 x 值。

$\begin{aligned} (g \circ f) (x) = g (f (x)) = g (\sqrt{x - 8}) = \frac{1}{2} [\sqrt{x - 8}]^{2} = \frac{1}{2} (x - 8) = \frac{1}{2} x - 4 \end{aligned}$

:g)f(x)=g(f(x))=g(x-8)=12[x-8]2=12(x-8)=12(x-8)=12x-4

Notice that $\begin{aligned} f (g (x)) \neq g (f (x)) \end{aligned}$ .
::通知 f( g( x))\\\ g( f( x) ) 。

by CK-12 demonstrates how to compose functions.
::CK-12 演示如何组成函数。

Finding the domain of a composition of functions can be tricky. Think about the process we follow to find a composition. First, we input one function into another. If the input function is undefined for some values of x , we will not be able to find $\begin{aligned} f (undefined) \end{aligned}$ . So, one of the things we need to consider in finding the domain of a composition is the domain of the input function.
::查找函数构成的域可能比较棘手。想想我们找到组成的过程。首先, 我们将一个函数输入到另一个函数中。如果输入函数没有定义 x 的某些值, 我们就无法找到 f( 未定义 ) 。因此, 在查找组成域时我们需要考虑的事情之一是输入函数的域。

We also need to consider the domain of the final result. Maybe by composing two functions that created a domain problem. In the domain of the composition, that should be taken into account as well.
::我们还需要考虑最终结果的领域,也许通过建立造成领域问题的两个功能,在构成领域,这也应该考虑。

Example 2
::例2

Find $\begin{aligned} (g \circ f) (x) \end{aligned}$ and $\begin{aligned} (f \circ f) (x) \end{aligned}$ when $\begin{aligned} g (x) = 4 x + 7 \end{aligned}$ and $\begin{aligned} f (x) = 5 x^{- 1} \end{aligned}$ and the domain of the compositions.
::当 g( x) =4x+7 和 f( x) = 5x- 1 和构成域时, 查找 (gf) (x) 和 (ff) (x) 。

Solution: a. First, let's find the composition $\begin{aligned} (g \circ f) (x) \end{aligned}$ . Plug $\begin{aligned} f (x) \end{aligned}$ into $\begin{aligned} g (x) \end{aligned}$ everywhere there is an $\begin{aligned} x \end{aligned}$ .
::解答 : a. 首先, 让我们找到组成成分 (gf(x) ) 。任何地方的f(x) 到 g(x) 中都有 x 。

$\begin{aligned} (g \circ f) (x) & = g (f (x)) \\ = g (5 x^{- 1}) \\ = 4 (5 x^{- 1}) + 7 \\ = 20 x^{- 1} + 7 o r \frac{20 + 7 x}{x} \end{aligned}$

:g*f(x)=g(f(x))=g(5x-1)=g(5x-1)=4(5x-1)+7=20x-1+7或20+7xx

The domain of the input function, $\begin{aligned} f (x) \end{aligned}$ , is all real numbers except $\begin{aligned} x \neq 0 \end{aligned}$ , because we cannot divide by zero. Therefore , the domain of $\begin{aligned} (g \circ f) (x) \end{aligned}$ is all real numbers except $\begin{aligned} x \neq 0 \end{aligned}$ .
::输入函数 f( x) 的域是除 x {% 0 以外的所有真实数字, 因为我们不能除以 0 。因此, (g }f (x) 的域是除 x {% 0 外的所有真实数字。

b. $\begin{aligned} (f \circ f) (x) \end{aligned}$ is a composite function on itself. We will plug $\begin{aligned} f (x) \end{aligned}$ into $\begin{aligned} f (x) \end{aligned}$ everywhere there is an $\begin{aligned} x \end{aligned}$ .
::b. (ff)(x) 本身是一个复合函数。我们将在有x的任何地方将f(x)插入f(x)至f(x)。

$\begin{aligned} (f \circ f) (x) & = f (f (x)) \\ = f (5 x^{- 1}) \\ = 5 (5 x^{- 1})^{- 1} \\ = 5 \cdot 5^{- 1} x^{1} power rule of exponents \\ = x \end{aligned}$

:f* f(x) = f(f(x)) = f(5x) - 1) = 5(5x) - 1) - 1= 1= 5*5= 1= 1x1 power expronents=x

Even though the domain of $\begin{aligned} x \end{aligned}$ is all real numbers, t he domain of the input function, $\begin{aligned} f (x) \end{aligned}$ , is all real numbers except $\begin{aligned} x \neq 0 \end{aligned}$ . Therefore, the domain of $\begin{aligned} (f \circ f) (x) \end{aligned}$ is all real numbers except $\begin{aligned} x \neq 0 \end{aligned}$ .
::即使 x 的域是所有真实数字, 但输入函数 f( x) 的域除 x\\\ 0 外都是真实数字。因此, (f\\\ f (x) 的域除 x\ *0 外都是真实数字。

WARNING
::警告

The symbols for multiplication $\begin{aligned} \cdot \end{aligned}$ and composition $\begin{aligned} \circ \end{aligned}$ are similar. Be careful not to mix them up.

Example 3
::例3

If $\begin{aligned} f (x) = x^{4} - 1 \end{aligned}$ and $\begin{aligned} g (x) = 2 \sqrt[4]{x + 1} \end{aligned}$ , find $\begin{aligned} g \circ f \end{aligned}$ and the restrictions on the domain.
::如果 f( x) =x4 - 1 和 g( x) = 2x+14, 请找到 gf 和对域的限制。

Solution: Recall that $\begin{aligned} g \circ f \end{aligned}$ is another way of writing $\begin{aligned} g (f (x)) \end{aligned}$ . Let’s plug $\begin{aligned} f \end{aligned}$ into $\begin{aligned} g \end{aligned}$ .
::解答: 回顾 gf 是写 g(f) (x) 的另一种方式。让我们将 f 插入 g 。

$\begin{aligned} (g \circ f) (x) & = g (f (x)) \\ = g (x^{4} - 1) \\ = 2 \sqrt[4]{(x^{4} - 1) + 1} \\ = 2 \sqrt[4]{x^{4}} \\ = 2 | x | \end{aligned}$

:g*f(x)=g(f(x))=g(x4-1)=g(x4-1)=2(x4-1)+14=2x44=2**x*)

The final function, $\begin{aligned} (g \circ f) (x) \neq 2 x \end{aligned}$ because $\begin{aligned} x \end{aligned}$ is being raised to the $\begin{aligned} 4^{t h} \end{aligned}$ power, which will always yield a positive result . Therefore, even when $\begin{aligned} x \end{aligned}$ is negative, the result will be positive. For example, if $\begin{aligned} x = - 2 \end{aligned}$ , then $\begin{aligned} g \circ f = 2 \sqrt[4]{{(- 2)}^{4}} = 2 \cdot 2 = 4. \end{aligned}$ . An absolute value function has no restrictions on the domain.
::最后函数, (gf)(x) 2x , 因为 x 正在被提升到第四权力, 这总是会产生正结果。因此, 即使 x 是负, 结果也会是正结果。例如, 如果 x2, 则 gf=2(- 2) 44= 22=4. 。绝对值函数对域没有限制。

Recall the previous example, however. The restrictions, if there are any, from the inner function, $\begin{aligned} f (x) \end{aligned}$ , still exist. Because there are no restrictions on $\begin{aligned} f (x) \end{aligned}$ , the domain of $\begin{aligned} g \circ f \end{aligned}$ remains all real numbers.
::不过,回顾上一个例子。内函数 f( x) 的限制仍然存在, 如果有的话, 内部函数 f( x) 的限制也依然存在。由于 f( x) 没有限制, gf 的域仍然是所有实际数字。

by CK-12 demonstrates how to find the composition of two functions.
::CK-12表明如何找到两个职能的组成。

Example 4
::例4

The number of bacteria in a petri dish in a lab can be modeled by $\begin{aligned} N (f) = \frac{- 100}{f - 2} + 50 \end{aligned}$ , where f is the temperature in Fahrenheit. The temperature in the lab can be modeled by $\begin{aligned} F (t) = - 2 t^{2} + 8 t + 65 \end{aligned}$ for $\begin{aligned} 0 \leq t \leq 4 \end{aligned}$ where t is time in hours. Find $\begin{aligned} N (F (t)) \end{aligned}$ to have a function of the number of bacteria over time.
::实验室中小盘中的细菌数量可以由 N(f) 100f-2+50 模拟, 其中 f 是 Fahrenheit 的温度。实验室中的温度可以由 F( t) 2t2+8t+65 模拟, 其值为 0. t4 , 其值为小时。查找 N( F( t) ) 以随时间推算细菌数量的函数。

Solution:
::解决方案 :

$\begin{aligned} N (F (t)) & = N (- 2 t^{2} + 8 t + 65) \\ = \frac{- 100}{(- 2 t^{2} + 8 t + 65) - 2} + 50 \\ = \frac{- 100}{- 2 t^{2} + 8 t + 63} + 50 \end{aligned}$

::N(F(t))=N(- 2t2+8t+65)\_100(- 2t2+8t+65)-2+50_100-2+2+100-2t2+8t+63+50

Since $\begin{aligned} F (t) \end{aligned}$ is defined only when $\begin{aligned} 0 \leq t \leq 4 \end{aligned}$ , the that holds true for $\begin{aligned} N (F (t)) \end{aligned}$ as well.
::由于F(t)的定义只在 0t4时才确定,因此对N(F(t))也适用。

Summary
::摘要

To compose two functions, input one function into the other function.
::要组成两个函数,则将一个函数输入到另一个函数中。

The domain of compositions is the intersection of the domain of the input function and the domain of the result of the composition.
::构成域是输入函数域与构成结果域的交叉点。

Review
::回顾

For problems 1-5, use the following functions to perform the following operations.
::对于问题1-5,使用下列功能来进行下列作业:

$\begin{aligned} f (x) = x^{2} + 5 g (x) = 3 \sqrt{x - 5} h (x) = 5 x + 1 \end{aligned}$

:x)=x2+5g(x)=3x-5h(x)=5x+1

1. $\begin{aligned} f \circ g \end{aligned}$
::1. fg

2. $\begin{aligned} h (f (x)) \end{aligned}$
::2.h(f(xx)))

3. $\begin{aligned} g \circ f \end{aligned}$
::3. gf

4. $\begin{aligned} g (h (x)) \end{aligned}$
::4.g(h(xx))

5. $\begin{aligned} f \circ g \circ h \end{aligned}$
::5. fghh

For problems 5-10, use the following functions to form the indicated compositions and clearly indicate any restrictions to the domain of the composite function.
::对于问题5-10,使用下列功能组成所示组成,并明确表明对复合功能领域的任何限制。

$\begin{aligned} p (x) = \frac{5}{x} q (x) = 5 \sqrt{x} r (x) = \frac{\sqrt{x}}{5} s (x) = \frac{1}{5} x^{2} \end{aligned}$

::p(x)=5xq(x)=5xr(x)=x5s(x)=15x2

6. $\begin{aligned} p (q (x)) \end{aligned}$
::6. p(q(xx))

7. $\begin{aligned} s (q (x)) \end{aligned}$
::7. s(q(x))

8. $\begin{aligned} q \circ s \end{aligned}$
::8. qs

9. $\begin{aligned} q \circ p \circ s \end{aligned}$
::9. qps

10. $\begin{aligned} p \circ r \end{aligned}$
::10. pr

Explore More
::探索更多

1. You go to a store to buy sneakers. You have a coupon for $10 off. There is also a sale for 30% off your purchase. Write two functions that model these discounts on an item that costs d dollars. Then, compose them both ways and determine if the order the discounts are applied in matters.
::1. 去商店买运动鞋。你有一个10美元的优惠券,还有30%的优惠券。请写两个功能来模拟这些折价对于一个费用为d美元的项目。然后,用两种方式组成它们,并确定这些折价单是否适用于事务。

2. The formula for an amount of money A where interest is compounded once a year is $\begin{aligned} A (P) = P (1 + r) \end{aligned}$ where P is the principal or the amount invested and r is the annual interest rate. Find a formula for the amount after 3 years by finding $\begin{aligned} A \circ A \circ A \end{aligned}$ .
::2. 利息每年复利一次的金额A的公式是A(P)=P(1+r),P为本金或投资额,r为年利率。在3年后,通过找到AAA找到金额的公式。

3. The radius of a ball changes according to the formula $\begin{aligned} r (t) = 4 t \end{aligned}$ , where t is in minutes and r is in centimeters. Find a model for the volume of air in the ball over time if the volume of a sphere is $\begin{aligned} V = \frac{4}{3} π r^{3} \end{aligned}$ .
::3. 根据公式r(t)=4t改变球的半径,t以分钟计,r以厘米计。如果球体的体积是V=43°r3,则寻找一个模型,以显示球体在一段时间内的空气量。

Answers to Review and Explore More Problems
::对审查和探讨更多问题的答复

Please see the Appendix.
::请参看附录。

PLIX
::PLIX

Try this interactive that reinforces the concepts explored in this section:
::尝试这一互动,强化本节所探讨的概念:

章节大纲

常规

职能的组成

Composition of Functions ::职能的组成

Example 1 ::例1

Example 2 ::例2

WARNING ::警告

Example 3 ::例3

Example 4 ::例4

Summary ::摘要

Review ::回顾

Explore More ::探索更多

Answers to Review and Explore More Problems ::对审查和探讨更多问题的答复

PLIX ::PLIX

Composition of Functions
::职能的组成

Example 1
::例1

Example 2
::例2

WARNING
::警告

Example 3
::例3

Example 4
::例4

Summary
::摘要

Review
::回顾

Explore More
::探索更多

Answers to Review and Explore More Problems
::对审查和探讨更多问题的答复

PLIX
::PLIX