课程： 7.10 职能业务 | The Way To Learn

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选择节职能业务

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职能业务
The area of a rectangle is $\begin{aligned} 2 x^{2} \end{aligned}$ . The length of the rectangle is $\begin{aligned} \sqrt{x + 3} \end{aligned}$ . What is the width of the rectangle? What restrictions, if any, are on this value?
::矩形区域是 2x2 。矩形的长度是 x+3 。矩形的宽度是多少? 此值上有什么限制?

Function Operations
::职能业务

We have already dealt with adding, subtracting, and multiplying functions. To add and subtract, you combine like terms . When multiplying, you either FOIL or use the “box” method. When you add, subtract, or multiply functions, it is exactly the same as what you would do with , except for the notation. Note that we don't need to write out the entire function $\begin{aligned} f (x) - g (x) \end{aligned}$ , just $\begin{aligned} f - g \end{aligned}$ , for example. Let’s continue:
::我们已经处理过添加、减法和乘法的问题。要添加和减法,您可以将术语合并起来。当乘法时,您要么是FOIL,要么是使用“框”方法。当您添加、减法或乘法时,它与您要使用的功能完全相同,但符号除外。请注意,我们不需要写出整个函数f(x)-g(x),只是f-g(x),例如f-g。让我们继续:

$\begin{aligned} f (x) - g (x) & = (x + 5) - (x^{2} - 4 x + 8) \\ = x + 5 - x^{2} + 4 x - 8 \\ = - x^{2} + 5 x - 3 \end{aligned}$

:x)-g(x)=(x+5)-(x2-4x+8)=x+5-x2+4x-8-8*x2+5x-3

Distribute the negative sign to the second function and combine like terms. Be careful! $\begin{aligned} f (x) - g (x) \neq g (x) - f (x) \end{aligned}$ . Also, this new function, $\begin{aligned} f (x) - g (x) \end{aligned}$ has a different domain and range that either $\begin{aligned} f (x) \end{aligned}$ or $\begin{aligned} g (x) \end{aligned}$ .
::将负符号分布到第二个函数中, 并合并类似术语。注意 ! f( x)- g( x)- g( x)- g( x)- f( x) 。另外, f( x)- g( x) 的新函数, f( f)- g( x) 具有不同的域和范围, 范围为 f( x) 或 g( x) 。

If $\begin{aligned} f (x) = \sqrt{x - 8} \end{aligned}$ and $\begin{aligned} g (x) = \frac{1}{2} x^{2} \end{aligned}$ , let's find $\begin{aligned} f g \end{aligned}$ and $\begin{aligned} \frac{f}{g} \end{aligned}$ and determine any restrictions for $\begin{aligned} \frac{f}{g} \end{aligned}$ .
::如果 f( x) =x- 8 和 g( x) = 12x2, 让我们找到 fg 和 fg 并确定对 fg 的限制。

First, even though the $\begin{aligned} x \end{aligned}$ is not written along with the $\begin{aligned} f (x) \end{aligned}$ and $\begin{aligned} g (x) \end{aligned}$ , it can be implied that $\begin{aligned} f \end{aligned}$ and $\begin{aligned} g \end{aligned}$ represent $\begin{aligned} f (x) \end{aligned}$ and $\begin{aligned} g (x) \end{aligned}$ .
::首先,即使x没有与 f(x)和 g(x)一起写,但可以暗示f和g代表f(x)和g(x)。

$\begin{aligned} f g = \sqrt{x - 8} \cdot \frac{1}{2} x^{2} = \frac{1}{2} x^{2} \sqrt{x - 8} \end{aligned}$
::fg=x- 8=12x2=12x2x- 8

To divide the two functions, we will place $\begin{aligned} f \end{aligned}$ over $\begin{aligned} g \end{aligned}$ in a fraction .
::为了将这两个函数分开,我们将把f 大于 g 分化成一个分数。

$\begin{aligned} \frac{f}{g} = \frac{\sqrt{x - 8}}{\frac{1}{2} x^{2}} = \frac{2 \sqrt{x - 8}}{x^{2}} \end{aligned}$
::fg=x- 812x2=2x-8x2

To find the restriction(s) on this function, we need to determine what value(s) of $\begin{aligned} x \end{aligned}$ make the denominator zero because we cannot divide by zero. In this case $\begin{aligned} x \neq 0 \end{aligned}$ . Also, the domain of $\begin{aligned} f (x) \end{aligned}$ is only $\begin{aligned} x \geq 8 \end{aligned}$ , because we cannot take the square root of a negative number. The portion of the domain where $\begin{aligned} f (x) \end{aligned}$ is not defined is also considered part of the restriction. Whenever there is a restriction on a function, list it next to the function, separated by a semi-colon. We will not write $\begin{aligned} x \neq 0 \end{aligned}$ separately because it is included in $\begin{aligned} x < 8 \end{aligned}$ .
::要找到此函数的限制, 我们需要确定 x 的值使分母为零, 因为我们不能除以零。在此情况下, x0。另外, f( x) 的域仅是 x8 , 因为我们无法从负数的平方根中取出负数。未定义 f( x) 的域部分也被视为限制的一部分。当函数有限制时, 请在函数旁边列出, 以分号分隔。我们不会单独写入 x0 , 因为它包含在 x<8 中。

$\begin{aligned} \frac{f}{g} = \frac{2 \sqrt{x - 8}}{x^{2}}; x < 8 \end{aligned}$
::fg=2x-8x2; x < 8

Now we will introduce a new way to manipulate functions; composing them. When you compose two functions, we put one function into the other, where ever there is an $\begin{aligned} x \end{aligned}$ . The notation can look like $\begin{aligned} f (g (x)) \end{aligned}$ or $\begin{aligned} f \circ g \end{aligned}$ , and is read “ $\begin{aligned} f \end{aligned}$ of $\begin{aligned} g \end{aligned}$ of $\begin{aligned} x \end{aligned}$ ”.
::现在,我们将引入一种新的操作函数的方法; 组合函数。当您组成两个函数时, 我们将一个函数放入另一个函数, 在那里永远有一个 x。标记可以看起来像 f( g( x)) 或 fg, 并且读作“ f of g of x ” 。

Using $\begin{aligned} f (x) \end{aligned}$ and $\begin{aligned} g (x) \end{aligned}$ from the previous problem, let's find $\begin{aligned} f (g (x)) \end{aligned}$ and $\begin{aligned} g (f (x)) \end{aligned}$ and any restrictions on the domains.
::使用上一个问题的 f( x) 和 g( x) , 让我们找到 f( g( x) ) 和 g( f( x) ) 以及对域的限制。

For $\begin{aligned} f (g (x)) \end{aligned}$ , 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.
::对于 f( g( x) ), 我们要将 g( x) 放在 f( x) 中, 任何地方都有 x 值。

$\begin{aligned} f (g (x)) = \sqrt{g (x) - 8} \end{aligned}$
:g(x))=g(x)-8

Now, substitute in the actual function for $\begin{aligned} g (x) \end{aligned}$ .
::现在,将实际函数中的 g(x) 替换为 g(x) 。

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

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

To find the domain of $\begin{aligned} f (g (x)) \end{aligned}$ , let’s determine where $\begin{aligned} x \end{aligned}$ is defined. The radicand is equal to zero when $\begin{aligned} x = 4 \end{aligned}$ or $\begin{aligned} x = - 4 \end{aligned}$ . Between 4 and -4, the function is not defined because the square root would be negative. Therefore , the domain is all real numbers; $\begin{aligned} - 4 < x < 4 \end{aligned}$ .
::要找到 f( g( x) 的域, 让我们来决定 x 定义的位置。当 x= 4 或 x\\ 4 时, radicand 等于 0 。在 4 和 4 之间, 函数没有定义, 因为平方根是负的。因此, 域是所有实际数字 ; - 4 < x < 4 。

Now, to find $\begin{aligned} g (f (x)) \end{aligned}$ , 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,要找到 g( f( x) ), 我们会把 f( x) 放到 g( x) 中, 任何地方都有 x 值。

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

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

Notice that $\begin{aligned} f (g (x)) \neq g (f (x)) \end{aligned}$ . It is possible that $\begin{aligned} f \circ g = g \circ f \end{aligned}$ and is a special case, which will be addressed later. To find the domain of $\begin{aligned} g (f (x)) \end{aligned}$ , we will determine where $\begin{aligned} x \end{aligned}$ is defined. $\begin{aligned} g (f (x)) \end{aligned}$ is a line, so we would think that the domain is all real numbers. However, while simplifying the composition, the square and square root canceled out. Therefore, any restriction on $\begin{aligned} f (x) \end{aligned}$ or $\begin{aligned} g (x) \end{aligned}$ would still exist. The domain would be all real numbers such that $\begin{aligned} x \geq 8 \end{aligned}$ from the domain of $\begin{aligned} f (x) \end{aligned}$ . Whenever operations cancel, the original restrictions from the inner function still exist. As with the case of $\begin{aligned} f (g (x)) \end{aligned}$ , no simplifying occurred, so the domain was unique to that function.
::注意 f( g( x) )\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\( g( g( g)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\F( g( g( g)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

If $\begin{aligned} f (x) = x^{4} - 1 \end{aligned}$ and $\begin{aligned} g (x) = 2 \sqrt[4]{x + 1} \end{aligned}$ , let's 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, 我们来看看 gf 和对域的限制。

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 $\begin{aligned} g \end{aligned}$ .
::回顾 gf 是写 g(f) (x) 的另一种方式。让我们将 f 插入 g 。

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

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

The final function, $\begin{aligned} g \circ f \neq 2 x \end{aligned}$ 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 answer. Therefore, even when $\begin{aligned} x \end{aligned}$ is negative, the answer 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. This will always happen when even roots and powers cancel. The range of this function is going to be all positive real numbers because the absolute value is never negative.
::最后函数, gf%2x 因为 x 正在被提升到第四权力区, 它总是产生一个正答案。因此, 即使 x 是负的, 答案也会是正的。例如, 如果 x2, 则gf=2( 2- 2) 44= 22=4. 。绝对值函数对域没有限制。即使在根和权力区取消时, 也总是会发生这种情况。此函数的范围将全部是正的, 因为绝对值从不负。

Recall, from the previous problem , however, that 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.
::然而,从前一个问题来看,如果有来自内函数f(x)的任何限制,f(x)仍然存在。由于f(x)没有限制,g(f)的域仍然是所有实际数字。

Examples
::实例

Example 1
::例1

Earlier, you were asked to determine if there are any restrictions on the width of the rectangle if the area is $\begin{aligned} 2 x^{2} \end{aligned}$ and the length is $\begin{aligned} \sqrt{x + 3} \end{aligned}$ .
::早些时候,有人要求你确定矩形宽度是否有限制,如果区域是 2x2,长度是 x+3 。

If we set g equal to $\begin{aligned} 2 x^{2} \end{aligned}$ and f equal to $\begin{aligned} \sqrt{x + 3} \end{aligned}$ , to find the width, we need to find $\begin{aligned} \frac{g}{f} \end{aligned}$ .
::如果我们设定 g 等于 2x2 和 f 等于 x+3 以找到宽度, 我们需要找到 gf 。

$\begin{aligned} \frac{2 x^{2}}{\sqrt{x + 3}} \\ \frac{2 x^{2}}{\sqrt{x + 3}} \cdot \frac{\sqrt{x + 3}}{\sqrt{x + 3}} \\ \frac{2 x^{2} \sqrt{x + 3}}{x + 3} \end{aligned}$

::2x2x+32x2x+3xx+3xx+3x+32x2x+3x+3

Therefore the width of the rectangle is $\begin{aligned} \frac{2 x^{2} \sqrt{x + 3}}{x + 3} \end{aligned}$ , and $\begin{aligned} x = - 3 \end{aligned}$ is a restriction on the answer.
::因此矩形的宽度为 2x2x+3x+3, x3 对答案是一个限制。

For the following examples, use $\begin{aligned} f (x) = 5 x^{- 1} \end{aligned}$ and $\begin{aligned} g (x) = 4 x + 7 \end{aligned}$ .
::以下示例使用 f(x)=5x-1和 g(x)=4x+7。

Example 2
::例2

Find $\begin{aligned} f g \end{aligned}$ .
::查找 fg.

$\begin{aligned} f g \end{aligned}$ is the product of $\begin{aligned} f (x) \end{aligned}$ and $\begin{aligned} g (x) \end{aligned}$ .
::fg 是 f(x) 和 g(x) 的产物。

$\begin{aligned} f g & = 5 x^{- 1} (4 x + 7) \\ = 20 x^{0} + 35 x^{- 1} \\ = 20 + 35 x^{- 1} o r \frac{20 x + 35}{x} \end{aligned}$

::fg=5x-1(4x+7)=20x0+35x-1=20+35x-1或20x+35x

Both representations are correct. Discuss with your teacher how s/he would like you to leave your answer.
::和老师讨论一下他希望你如何不回答你的问题

Example 3
::例3

Find $\begin{aligned} g - f \end{aligned}$ .
::寻找 g-f.

Subtract $\begin{aligned} f (x) \end{aligned}$ from $\begin{aligned} g (x) \end{aligned}$ and simplify, if possible.
::从 g(x) 中减法(x)并尽可能简化。

$\begin{aligned} g - f & = (4 x + 7) - 5 x^{- 1} \\ = 4 x + 7 - 5 x^{- 1} o r \frac{4 x^{2} + 7 x - 5}{x} \end{aligned}$

::g- f=( 4x+7)- 5x- 1=4x+7- 5x-1 或 4x2+7x- 5x

Example 4
::例4

Find $\begin{aligned} \frac{f}{g} \end{aligned}$ .
::查找 fg.

Divide $\begin{aligned} f (x) \end{aligned}$ by $\begin{aligned} g (x) \end{aligned}$ . Don’t forget to include the restriction(s).
::将f(x)除以 g(x) 。不要忘记包含限制。

$\begin{aligned} \frac{f}{g} & = \frac{5 x^{- 1}}{4 x + 7} \\ = \frac{5}{x (4 x + 7)}; x \neq 0, - \frac{7}{4} \end{aligned}$

::fg=5x-14x+7=5x(4x+7); x0,-74

Recall the properties of exponents. Anytime there is a negative exponent , it should be moved into the denominator. We set each factor in the denominator equal to zero to find the restrictions.
::回想指数的属性。每当出现负指数时, 它应该被移动到分母中。我们设置分母中的每个系数为零, 以找到限制。

Example 5
::例5

Find $\begin{aligned} g (f (x)) \end{aligned}$ and the domain.
::查找 g(f(x)) 和域名。

$\begin{aligned} g (f (x)) \end{aligned}$ is a composition function. Let’s plug $\begin{aligned} f (x) \end{aligned}$ into $\begin{aligned} g (x) \end{aligned}$ everywhere there is an $\begin{aligned} x \end{aligned}$ .
::g(f(x)) 是一个组成函数。让我们将f(x) 插入到 g(x) 中, 随处都有 x。

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

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

The domain of $\begin{aligned} f (x) \end{aligned}$ is all real numbers except $\begin{aligned} x \neq 0 \end{aligned}$ , because we cannot divide by zero. Therefore, the domain of $\begin{aligned} g (f (x)) \end{aligned}$ is all real numbers except $\begin{aligned} x \neq 0 \end{aligned}$ .
::f( x) 域是所有真实数字, 但 x* 0 除外, 因为我们不能除以 0 。因此, g( f( x) ) 域是除 x * 0 外的所有真实数字。

Example 6
::例6

Find $\begin{aligned} f \circ f \end{aligned}$ .
::寻找ff。

$\begin{aligned} f \circ f \end{aligned}$ 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 $\begin{aligned} x \end{aligned}$ .
::f*f 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) 。

$\begin{aligned} f (f (x)) & = 5 (f (x))^{- 1} \\ = 5 (5 x^{- 1})^{- 1} \\ = 5 \cdot 5^{- 1} x^{1} \\ = x \end{aligned}$

::f( f( x)) = 5( f( x)) - 1= 5( 5x) - 1) - 1= 5( 5x- 1) - 1= 5°5 - 1x1=x

Review
::回顾

For problems 1-8, 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1-8,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。

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

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

$\begin{aligned} f + h \end{aligned}$
::f+h 时

$\begin{aligned} h - g \end{aligned}$
::h-g

$\begin{aligned} \frac{f}{g} \end{aligned}$
::f克( fg)

$\begin{aligned} f h \end{aligned}$
::fh 时

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

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

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

$\begin{aligned} f \circ g \circ h \end{aligned}$
::{\fn方正黑体简体\fs18\b1\bord1\shad1\3cH2F2F2F

For problems 9-16, 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9-16,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

$\begin{aligned} p s \end{aligned}$
::p 年

$\begin{aligned} \frac{q}{r} \end{aligned}$
::qr

$\begin{aligned} q + r \end{aligned}$
::q+r +r

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

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

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

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

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

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

常规

职能业务

Function Operations ::职能业务

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)

Function Operations
::职能业务

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)