📚 Exercise 7E: Mastering Composite and Inverse Functions | 练习7E：掌握复合函数与反函数
Exercise 7E is a critical checkpoint in any IB Mathematics course, whether you are pursuing Analysis and Approaches or Applications and Interpretation. This exercise consolidates your understanding of how functions can be combined into composite functions and how to reverse a function's operation through its inverse. Mastery here is not just about algebraic manipulation; it demands a deep appreciation of domain, range, and the symmetrical relationship between a function and its inverse. In this article, we will break down the core concepts behind Exercise 7E, walk through typical problem types, and highlight the subtle pitfalls that often catch students off guard.
练习7E是任何IB数学课程中的一个关键节点，无论你学习的是分析与方法还是应用与解释。该练习巩固了你对如何将函数组合成复合函数以及如何通过反函数逆转函数运算的理解。掌握这部分内容不仅仅是代数操作，还需要深刻理解定义域、值域以及函数与其反函数之间的对称关系。在本文中，我们将分解练习7E背后的核心概念，讲解典型题型，并指出那些经常让学生措手不及的细微陷阱。
1. Understanding Functions as Mappings | 将函数理解为映射
ssigns exactly one output to each input from its domain. In IB terminology, we write f : x ↦ ermissible inputs is the domain, and the set of all outputs actually produced is the range. A firm grasp of set notation, including interval and inequality forms, is essential for dealing with the domain restrictions that frequently appear in composite and inverse function problems.
函数 f 是一种规则，它为定义域内的每个输入分配恰好一个输出。在IB术语中，我们写作 f : x ↦ f(x) 或简单地 y = f(x)。所有允许的输入构成的集合是定义域，而实际产生的所有输出构成的集合是值域。熟练掌握集合符号，包括区间和不等式形式，对于处理复合函数和反函数问题中经常出现的定义域限制至关重要。
For example, if f(x) = √x − 2 0. When a function is defined by a formula without any mention of domain, we assume the natural domain, i.e., all real numbers x for which the expression makes sense.
例如，如果 f(x) = √x − 2，那么定义域是 x ≥ 2，值域是 y ≥ 0。当函数由公式定义且未提及定义域时，我们采用自然定义域，即所有使表达式有意义的实数 x。
2. Composite Functions Defined | 复合函数的定义
The composite function (f ∘ g)(x) means we first apply g to x, and then apply f to the result. It is read as "f composed with g" or "f of g of x". The operation is only defined when the range of g overlaps with the domain of f sufficiently to produce valid outputs. Therefore, determining the domain of a composite function requires careful checking that g(x) belongs to the domain of f.
复合函数 (f ∘ g)(x) 表示我们先对 x 应用函数 g，然后再对结果应用函数 f。它读作“f 与 g 的复合”或“f of g of x”。只有当 g 的值域与 f 的定义域有足够交集以产生有效输出时，该运算才有定义。因此，确定复合函数的定义域需要仔细检查 g(x) 是否属于 f 的定义域。
Symbolically: (f ∘ composition is not commutative in general; f(g(x)) and g(f(x)) are usually different functions.
符号表示为：(f ∘ g)(x) = f(g(x))。特别注意，复合运算通常不交换；f(g(x)) 和 g(f(x)) 一般是不同的函数。
3. Notation and Domain Sketching for Composites | 复合函数的记法与定义域草图
x)), start with the inner function g. The x-values must first belong to the domain of g. Then, the output g(x) must lie within the domain of f. Both conditions must be satisfied simultaneously. Drawing a domain "funnel" diagram can be a powerful visual aid.
求 h(x) = f(g(x)) 的定义域时，从内层函数 g 开始。x 值必须首先属于 g 的定义域。然后，输出 g(x) 必须落在 f 的定义域内。这两个条件必须同时满足。绘制定义域“漏斗”图可以是一个强有力的视觉辅助工具。
For instance, if f(x) = √x nd g(x) = x² − 9, then the natural domain of g is all real numbers. However, for f(g(x)) = √x² − 9 to be defined, we require x² − 9 ≥ 0, so x ≤ −3 or x ≥ 3. Thus the domain of f ∘ g is {x ∈ ℝ : x ≤ −3 or x ≥ 3}.
例如，如果 f(x) = √x 且 g(x) = x² − 9，那么 g 的自然定义域是所有实数。但是，要使 f(g(x)) = √x² − 9 有定义，我们需要 x² − 9 ≥ 0，因此 x ≤ −3 或 x ≥ 3。所以 f ∘ g 的定义域是 {x ∈ ℝ : x ≤ −3 或 x ≥ 3}。
4. Algebraic Construction of Composite Functions | 复合函数的代数构建
To form (f ∘ g)(x), substitute the entire expression for g(x) into f(x). Be meticulous with brackets, especially when squaring, taking reciprocals, or dealing with logarithms. After substitution, always simplify if possible, but the simplified form does not automatically give the correct domain—the original restrictions remain.
要构建 (f ∘ g)(x)，将 g(x) 的完整表达式代入 f(x)。要格外注意括号，特别是在平方、取倒数或涉及对数时。代入后，尽可能简化，但简化后的形式并不会自动给出正确的定义域——原始的限制条件仍然保留。
Example: f(x) = 1/x, g(x) = x + 3 ⇒ (f ∘ g)(x) = 1/(x + 3), x ≠ −3.
例子：f(x) = 1/x, g(x) = x + 3 ⇒ (f ∘ g)(x) = 1/(x + 3), x ≠ −3。
5. Inverse Functions: Reversing the Operation | 反函数：逆转运算
o-one (injective) on its domain to possess an inverse. If it is not injective, the domain can sometimes be restricted to make it so. In IB, you often need to state the largest possible domain for which a function is invertible.
反函数 f⁻¹ 逆转了 f 的作用。正式地，如果 f(a) = b，那么 f⁻¹(b) = a。一个函数在其定义域上必须是一一映射（单射）才能拥有反函数。如果它不是单射，有时可以通过限制定义域使其成为单射。在IB中，你经常需要说明使函数可逆的最大可能定义域。
A common mistake is to confuse f⁻¹(x) with [f(x)]⁻¹ which denotes the reciprocal. The context makes it clear: f⁻¹ is the inverse function, not 1/f(x).
一个常见错误是将 f⁻¹(x) 与 [f(x)]⁻¹（表示倒数）混淆。上下文会明确区别：f⁻¹ 是反函数，而不是 1/f(x)。
6. Finding Inverse Functions Algebraically | 代数求反函数
hen solve for y. This new y is f⁻¹(x). Always express the inverse using function notation and state its domain, which is the range of the original function. The range of f⁻¹ will be the domain of the original function.
求 f⁻¹(x) 的标准方法从写出 y = f(x) 开始。交换 x 和 y，然后解出 y。这个新的 y 就是 f⁻¹(x)。始终用函数符号表示反函数，并说明其定义域，即原函数的值域。f⁻¹ 的值域将是原函数的定义域。
If f(x) = 2x − 4, then y = 2x − 4. Swap: x = 2y − 4 ⇒ y = (x + 4)/2. Thus f⁻¹(x) = (x + 4)/2.
如果 f(x) = 2x − 4，则 y = 2x − 4。交换：x = 2y − 4 ⇒ y = (x + 4)/2。因此 f⁻¹(x) = (x + 4)/2。
7. Domain and Range of Inverse Functions | 反函数的定义域与值域
The domain of f⁻¹ is exactly the range of f. If the domain of f has been restricted to make it invertible, this restriction directly affects the range of f⁻¹. You must calculate or deduce the range of the original function accurately—graphical methods or using calculus can help find maximums, minimums, and asymptotic behaviour.
f⁻¹ 的定义域恰好是 f 的值域。如果 f 的定义域被限制以使其可逆，这个限制会直接影响 f⁻¹ 的值域。你必须准确计算或推导原函数的值域——图形方法或使用微积分可以帮助找到最大值、最小值和渐近行为。
| Function | Domain | Range |
| f(x) = x², x ≥ 0 | x ≥ 0 | y ≥ 0 |
| f⁻¹(x) = √x | x ≥ 0 | y ≥ 0 |
Notice the perfect swap. Always verify that your algebraic inverse produces correct domain-range correspondence.
请注意完美的对调。始终验证你代数求得的反函数产生了正确的定义域-值域对应关系。
8. Composition of a Function and Its Inverse | 函数与其反函数的复合
on with its inverse yields the identity function, but only on the appropriate domain. (f⁻¹ ∘ f)(x) = x for all x in the domain of f, and (f ∘ in the domain of f⁻¹. This property is often used to verify inverses and solve equations.
最优雅的结果之一是，将函数与其反函数复合会得到恒等函数，但只在适当的定义域上成立。对于 f 定义域中的所有 x，有 (f⁻¹ ∘ f)(x) = x；对于 f⁻¹ 定义域中的所有 x，有 (f ∘ f⁻¹)(x) = x。这个性质常被用来验证反函数和求解方程。
However, be careful: f⁻¹(f(x)) simplifies to x only when x is in the domain where f is defined and the composition makes sense. Always state the domain of validity.
但要注意：只有当 x 在 f 有定义的区域且复合有意义时，f⁻¹(f(x)) 才能化简为 x。务必说明有效定义域。
9. Graphical Interpretation and Symmetry | 图像解释与对称性
ur algebraic inverse is reasonable, and it helps you find intersections with the line y = x, which are invariant points.
在图像上，反函数是原函数关于直线 y = x 的反射。这种反射交换了 x 和 y 坐标的角色。如果点 (a, b) 在 f 的图像上，那么 (b, a) 在 f⁻¹ 的图像上。在同一坐标系中绘制两者可以立刻揭示你代数求得的反函数是否合理，并帮助你找到与直线 y = x 的交点，这些点是不变点。
oints satisfy f(x) = x. For an invertible function, these remain unchanged by the inverse.
不变点满足 f(x) = x。对于一个可逆函数，这些点不会被反函数改变。
10. Common Pitfalls in Exercise 7E | 练习7E中的常见错误
g mistakes: forgetting to state the domain of a composite or inverse function; assuming (f ∘ g)(x) = (g ∘ correctly interpreting the notation f⁻¹(x) as a power. Another subtle error is simplifying √x² to x without considering absolute value, which matters when solving equations involving inverses of squared functions.
练习7E经常暴露这些反复出现的错误：忘记说明复合函数或反函数的定义域；假设 (f ∘ g)(x) = (g ∘ f)(x)；错误使用括号导致代数错误；在对二次函数求逆之前忽略限制其定义域；以及错误地将符号 f⁻¹(x) 解释为幂次。另一个细微错误是将 √x² 简化为 x 而不考虑绝对值，这在求解涉及平方函数反函数的方程时很重要。
Always double-check restrictions. If original f(x) = x², x ≤ 0, then f⁻¹(x) = −√x, not √x. The sign reflects the restricted domain.
务必反复检查限制条件。如果原函数 f(x) = x², x ≤ 0，那么 f⁻¹(x) = −√x，而不是 √x。符号反映了受限制的定义域。
11. Practice Problem Walkthrough | 典型习题演练
Consider f(x) = 2x + 1 and g(x) = x² − 4. Let us find (f ∘ g)(x) and its domain, (g ∘ f)(x), and determine the inverse of f. First, (f ∘ ed for all x, and the output x² − 4 is always in the domain of f (all reals), so domain is ℝ. Second, (g ∘ f)(x) = g(2x+1) = (2x+1)² − 4 = 4x² + 4x − 3. Domain is also ℝ. Clearly they are different. Inverse of f: y = 2x+1 ⇒ x = 2y+1 ⇒ y = (x−1)/2, so f⁻¹(x) = (x−1)/2. Domain ℝ, range ℝ.
考虑 f(x) = 2x + 1 和 g(x) = x² − 4。我们来求 (f ∘ g)(x) 及其定义域、(g ∘ f)(x)，并确定 f 的反函数。首先，(f ∘ g)(x) = f(g(x)) = 2(x² − 4) + 1 = 2x² − 7。定义域：g(x) 对所有 x 有定义，且输出 x² − 4 始终在 f 的定义域（所有实数）内，因此定义域是 ℝ。其次，(g ∘ f)(x) = g(2x+1) = (2x+1)² − 4 = 4x² + 4x − 3，定义域也是 ℝ。显然它们不同。f 的反函数：y = 2x+1 ⇒ x = 2y+1 ⇒ y = (x−1)/2，因此 f⁻¹(x) = (x−1)/2。定义域 ℝ，值域 ℝ。
king composition: f⁻¹(f(x)) = ((2x+1)−1)/2 = x, and f(f⁻¹(x)) = 2((x−1)/2)+1 = x. Perfect.
验证复合：f⁻¹(f(x)) = ((2x+1)−1)/2 = x，并且 f(f⁻¹(x)) = 2((x−1)/2)+1 = x。完美。
12. IB Exam Tips for Composite and Inverse Functions | IB考试关于复合与反函数的提示
e y = x and indicate invariant points. If a function is not one-to-one, you may be asked to suggest a suitable domain restriction, so justify your choice. Finally, practice simplifying expressions like f⁻¹(f(x)) with explicit domain constraints—examiners love testing these details.
在IB考试中，涉及练习7E相关主题的题目分值较重，并常常综合多个概念。务必仔细阅读题目是要 f(g(x)) 的定义域还是仅仅表达式。用集合符号或不等式清晰地标明定义域限制。画图时，标注直线 y = x 并指出不变点。如果函数不是一一映射，你可能需要提出合适的定义域限制，因此要说明理由。最后，练习在明确的定义域限制下化简像 f⁻¹(f(x)) 这样的表达式——考官喜欢测试这些细节。
Remember that correct notation and clear reasoning are just as important as the final answer. Practice from Exercise 7E until the logic becomes second nature.
记住，正确的符号和清晰的推理与最终答案同样重要。反复练习练习7E，直到其中的逻辑成为你的第二天性。
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图