函数、复合、变换与逆运算
ACT 知识点 11 3:48 英语旁白 · 英中内嵌字幕
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Composition applies the inside rule first.
复合运算先应用内部规则。
If f of u equals one over u minus 2, and g of x equals x squared plus 2, then f of g of x equals one over x squared, with x not equal to zero.
若f(u)等于1除以(u减2)，且g(x)等于x平方加2，则f(g(x))等于1除以x平方，其中x不等于0。
An inverse reverses input and output on a one-to-one domain.
反函数在一对一定义域上反转输入与输出。
Restrict h of x equals the quantity x minus 1 squared to x greater than or equal to 1 before taking h inverse of x equals 1 plus the square root of x.
在求h的反函数为1加根号x之前，需将h(x)等于(x减1)的平方限制在x大于等于1的范围内。
Each task gives its own data.
每个任务都有各自的数据。
Do not copy the worked example’s numbers.
请勿照搬例题中的数值。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
请利用前几页的完整背景信息。
f of g of x is one over x squared plus 3 minus 3, which is one over x squared, excluding x equals 0.
f(g(x)) 等于 x 平方加三减三分之一，即 x 平方分之一，并排除 x 等于零。
For the other order, f of 4 is 1, and g of 1 is 4.
换一个顺序，f(4) 等于一，g(1) 等于四。
Composition order changes both the operation and possibly the domain; f of g of 4 is one sixteenth, which is different.
复合的顺序既改变运算，也可能改变定义域；f(g(4)) 等于十六分之一，结果不同。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
使用前面页面的完整背景信息。
Solve y equals x plus 2 squared, with x plus 2 at most 0, to get x equals negative 2 minus root y.
在 x 加二小于等于零的条件下解 y 等于 x 加二的平方，得到 x 等于负二减根号 y。
So the inverse, h inverse of x, is negative 2 minus root x, with domain x at least 0 and range y at most negative 2.
所以反函数 h 逆(x) 等于负二减根号 x，定义域是 x 大于等于零，值域是 y 小于等于负二。
The plus branch returns values outside the chosen original domain.
取加号的那一支，得到的值不在原来选定的定义域内。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
请利用前几页的完整背景信息。
g of x equals 3 times x plus 4 squared, minus 2, has vertex negative 4 comma negative 2, and range y at least negative 2.
g(x) 等于三乘 x 加四的平方再减二，顶点是（负四，负二），值域是 y 大于等于负二。
The old point 1 comma 1 maps to negative 3 comma 1, because 3 times 1 minus 2 equals 1; substitution confirms g of negative 3 equals 1.
原来的点（一，一）映射到（负三，一），因为三乘一减二等于一；代入可验证 g(负三) 等于一。
Return to the task and explain why the method fits.
回到该任务，并解释为何此方法适用。
Original practice does not predict an official score.
原始练习无法预测正式成绩。
Complete required live or recorded evidence separately.
请单独完成所需的现场或录制证据。
Pause and test yourself. Cover the worked answer, rebuild the answer to the last example in your own words, and name one condition that would change it before you check.
暂停一下，自我检测：遮住示范答案，用自己的话重做最后一个例子的答案，并在核对之前说出一个会改变结论的条件。