Composition of Functions · 函数的复合
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| composition/ˌkɒmpəˈzɪʃn/ | 复合 | fù hé |
| composite function/ˈkɒmpəzɪt ˈfʌŋkʃn/ | 复合函数 | fù hé hán shù |
| domain/dəˈmeɪn/ | 定义域 | dìng yì yù |
Functions in a chain
- Feed a number into one function, then feed its answer into a second function.
- The two machines work in a row: output of the first becomes input of the second.
- This chaining is called composition, and it builds complex functions from simple ones.
- It is how a "convert then round" or "discount then tax" calculation really works.
串成一条链的函数
- 把一个数送进一个函数,再把它的答案送进第二个函数。
- 两台机器一前一后工作:第一个的输出成为第二个的输入。
- 这种串联叫做复合,它用简单的函数搭建出复杂的函数。
- "先换算再取整"或"先打折再计税"的计算,实际上就是这样运作的。
The composite function
- The composite function 复合函数 is written $(f \circ g)(x) = f(g(x))$.
- Read it inside-out: apply $g$ first, then apply $f$ to the result.
- The composition 复合 hands the output of $g$ straight to $f$ as its input.
- So $g$ is the inner function and $f$ is the outer one.
复合函数
- 复合函数(composite function)写作 $(f \circ g)(x) = f(g(x))$。
- 从里到外读:先作用 $g$,再把结果作用 $f$。
- 复合(composition)把 $g$ 的输出直接交给 $f$ 作为它的输入。
- 所以 $g$ 是内层函数,$f$ 是外层函数。
Send an input through two functions in a row · 让一个输入依次经过两个函数
Step through the pipeline — the output of g is handed straight to f. · 一步步走过这条流水线——g 的输出被直接交给 f。
In the composition $(f \circ g)(x) = f(g(x))$, which function acts first · 第一个? · 在复合 $(f \circ g)(x) = f(g(x))$ 中,哪个函数先作用?
Work inside-out: $g$ acts on $x$ first, then $f$ acts on the result $g(x)$. · 从里到外:$g$ 先作用于 $x$,然后 $f$ 作用于结果 $g(x)$。
Evaluating a composition
- Always work from the inside out.
- To find $f(g(2))$, first compute $g(2)$, then feed that number into $f$.
- With $g(x)=x+1$ and $f(x)=x^2$: $g(2)=3$, then $f(3)=9$.
- You can also compose the formulas to get one rule for the whole chain.
求复合的值
- 永远从里到外算。
- 要求 $f(g(2))$,先算 $g(2)$,再把那个数送进 $f$。
- 当 $g(x)=x+1$、$f(x)=x^2$:$g(2)=3$,再 $f(3)=9$。
- 你也可以直接复合公式,得到整条链的一个规则。
If $g(x) = x + 1$ and $f(x) = x^2$, what is $f(g(2))$? · 若 $g(x) = x + 1$ 且 $f(x) = x^2$,$f(g(2))$ 是多少?
Inner first: $g(2) = 3$. Then outer: $f(3) = 3^2 = 9$. · 先算内层:$g(2) = 3$。再算外层:$f(3) = 3^2 = 9$。
The domain of a composition
- The composite is only defined where the chain actually works.
- The value $g(x)$ must be a legal input for $f$ — it must lie in $f$'s domain 定义域.
- If $g$ produces a number $f$ cannot accept, that $x$ is excluded.
- So the composite's domain can be smaller than $g$'s alone.
复合的定义域
- 只有在这条链真正行得通的地方,复合才有定义。
- $g(x)$ 的值必须是 $f$ 的合法输入——它必须落在 $f$ 的定义域(domain)内。
- 如果 $g$ 产生一个 $f$ 无法接受的数,那个 $x$ 就被排除。
- 所以复合的定义域可能比 $g$ 单独的更小。
For · 支持 $g(x)$ to be usable inside $f$, the value $g(x)$ must lie in the ____ of $f$. · 要让 $g(x)$ 能被 $f$ 使用,$g(x)$ 的值必须落在 $f$ 的____内。
The composite is defined only where $g(x)$ is a legal input to $f$ — inside $f$'s domain. · 只有当 $g(x)$ 是 $f$ 的合法输入时——即在 $f$ 的定义域内——复合才有定义。
Select all · 所有 true statements about composition. · 选出关于复合的所有正确说法。
Order usually does · 作用 matter — composition is not commutative. The other three are correct. · 顺序通常确实重要——复合不可交换。其余三条正确。
Order matters
- Composition is generally not commutative: $f(g(x)) \neq g(f(x))$.
- "Put on socks, then shoes" is not the same as "shoes, then socks".
- With $f(x)=x^2$, $g(x)=x+1$: $f(g(x))=(x+1)^2$ but $g(f(x))=x^2+1$.
- Always respect which function is inner and which is outer.
顺序很重要
- 复合通常不可交换:$f(g(x)) \neq g(f(x))$。
- "先穿袜子,再穿鞋"和"先穿鞋,再穿袜子"不一样。
- 当 $f(x)=x^2$、$g(x)=x+1$:$f(g(x))=(x+1)^2$,而 $g(f(x))=x^2+1$。
- 永远要分清哪个是内层,哪个是外层。
Composition is commutative: $f(g(x))$ always equals $g(f(x))$. · 复合是可交换的:$f(g(x))$ 总是等于 $g(f(x))$。
Order matters. With $f(x)=x^2$ and $g(x)=x+1$: $f(g(x))=(x+1)^2$ but $g(f(x))=x^2+1$ — different. · 顺序很重要。当 $f(x)=x^2$、$g(x)=x+1$:$f(g(x))=(x+1)^2$,而 $g(f(x))=x^2+1$——不一样。
Do not confuse $f(g(x))$ with $f(x)\cdot g(x)$. Composition feeds one function into another; it is not multiplication. And it rarely commutes, so swapping the order usually changes the answer.
不要把 $f(g(x))$ 与 $f(x)\cdot g(x)$ 搞混。复合是把一个函数送进另一个;它不是乘法。而且它很少可交换,所以交换顺序通常会改变答案。
Let $g(x) = \sqrt{x}$ and $f(x) = x + 3$. Find $(f \circ g)(x)$ and its domain.
- Inner first: $g(x) = \sqrt{x}$, then outer: $f(\sqrt{x}) = \sqrt{x} + 3$.
- The square root needs $x \ge 0$, so the domain is $x \ge 0$.
- Even though $f$ alone accepts all reals, the inner $g$ restricts the composite.
设 $g(x) = \sqrt{x}$、$f(x) = x + 3$。求 $(f \circ g)(x)$ 及其定义域。
- 先内层:$g(x) = \sqrt{x}$,再外层:$f(\sqrt{x}) = \sqrt{x} + 3$。
- 平方根需要 $x \ge 0$,所以定义域是 $x \ge 0$。
- 尽管 $f$ 单独接受所有实数,内层的 $g$ 却限制了复合。
A composite function $(f \circ g)(x) = f(g(x))$ chains two functions: the inner $g$ acts first, then $f$. Evaluate inside-out, and remember its domain needs $g(x)$ to be a legal input for $f$. Composition is generally not commutative — order matters.
复合函数 $(f \circ g)(x) = f(g(x))$ 把两个函数串起来:内层 $g$ 先作用,然后 $f$。从里到外求值,并记住它的定义域要求 $g(x)$ 是 $f$ 的合法输入。复合通常不可交换——顺序很重要。