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IB Mathematics: Functions Concepts & Key Exam Points | IB数学:函数概念与考点精讲

2026-08-30 16:52 · 本文部分内容由AI辅助创作

📚 IB Mathematics: Functions Concepts & Key Exam Points | IB数学:函数概念与考点精讲

Functions are the backbone of IB Mathematics. Whether you are taking Analysis and Approaches (AA) or Applications and Interpretation (AI), a deep understanding of function notation, domain and range, transformations, and inverses is essential for exam success. This article breaks down the core ideas and highlights the most common exam traps.

函数是IB数学的核心内容。无论你学习的是分析与方法(AA)还是应用与解释(AI),深刻理解函数记号、定义域与值域、图像变换以及反函数,都是取得考试成功的关键。这篇文章将拆解核心概念,并指出最常见的考试陷阱。

1. What is a Function? | 什么是函数?

ere A is the domain and B is the codomain. The set of all outputs actually produced is called the range.

函数是一种关系:每个输入(通常为 x)恰好对应一个输出(通常为 y)。我们写作 f: A → B,其中 A 是定义域,B 是陪域;实际产生的所有输出组成的集合称为值域。


2. Domain and Range | 定义域与值域

The domain of a function is the set of all possible input values for which the function is defined. The range is the set of resulting output values. In IB exams you must actively exclude values that cause division by zero, square roots of negative numbers, or logs of non-positive numbers.

函数的定义域是所有使函数有意义的输入值的集合;值域是相应输出值的集合。在IB考试中,你必须主动排除导致分母为零、偶次根号下为负或对数自变量非正的值。


3. Composite Functions | 复合函数

A composite function applies one function after another. (f ∘ g)(x) = f(g(x)) means first apply g to x, then apply f to the result. The domain of f ∘ satisfy two conditions: x must be in the domain of g, and g(x) must be in the domain of f.

复合函数是一个接一个地应用函数。(f ∘ g)(x) = f(g(x)) 表示先对 x 应用 g,再对结果应用 f。f ∘ g 的定义域必须满足两个条件:x 在 g 的定义域内,且 g(x) 在 f 的定义域内。

For example, if f(x) = x² and g(x) = x + 1, then (f ∘ g)(x) = f(g(x)) = (x + 1)². Order matters: (g ∘ f)(x) = g(x²) = x² + 1, which is different.

例如,若 f(x) = x²,g(x) = x + 1,则 (f ∘ g)(x) = f(g(x)) = (x + 1)²。顺序很重要:(g ∘ f)(x) = g(x²) = x² + 1,显然不同。


4. Inverse Functions | 反函数

y = f(x), then solve for y. The graph of f⁻¹ is the reflection of f across the line y = x.

反函数反转原函数的作用。只有当原函数是一一对应(单射)时,反函数才存在,即通过水平直线检验。求反函数的方法:在 y = f(x) 中将 x 与 y 互换,然后解出 y。f⁻¹ 的图像是 f 关于直线 y = x 的反射。

= 2x + 3, swap to get x = 2y + 3, and solve: y = (x - 3)/2. Therefore f⁻¹(x) = (x - 3)/2.

例如,若 f(x) = 2x + 3,先写成 y = 2x + 3,互换得 x = 2y + 3,再解出 y:y = (x - 3)/2,所以 f⁻¹(x) = (x - 3)/2。


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5. Transformations of Graphs | 函数图像的变换

You must be able to describe translations, reflections, and stretches from the equation form. A translation moves the graph, a reflection flips it, and a stretch changes its scale.

你必须能够根据方程描述平移、反射和伸缩。平移移动图像,反射翻转图像,伸缩改变图像的比例。

Transformation (变换)Effect / Example (效果/示例)
y = f(x) + aVertical shift up by a (向上平移 a 个单位)
y = f(x) - aVertical shift down by a (向下平移 a 个单位)
y = f(x - a)Horizontal shift right by a (向右平移 a 个单位)
y = f(x + a)Horizontal shift left by a (向左平移 a 个单位)
y = -f(x)Reflection in the x-axis (关于 x 轴对称)
y = f(-x)Reflection in the y-axis (关于 y 轴对称)
y = a·f(x)Vertical stretch by factor a (纵向拉伸 a 倍)
y = f(ax)Horizontal compression/stretch by factor 1/a (横向压缩/拉伸 1/a 倍)

or instance, y = (x - 3)² + 2 is the graph of y = x² shifted 3 units right and 2 units up.

例如,y = (x - 3)² + 2 就是 y = x² 向右平移3个单位、向上平移2个单位后的图像。


6. Quadratic Functions | 二次函数

ertex is at x = -b/(2a), and the axis of symmetry is the vertical line through the vertex.

二次函数写作 f(x) = ax² + bx + c(a ≠ 0)。其图像是抛物线。顶点位于 x = -b/(2a),对称轴是过该顶点的竖直直线。

= a(x - h)² + k, the vertex is (h, k). The roots can be found using the quadratic formula:

在顶点式中,f(x) = a(x - h)² + k,顶点为 (h, k)。根可通过求根公式得到:

x = (-b ± √b² - 4ac) / (2a)

of the roots: if Δ > 0, two distinct real roots; if Δ = 0, one repeated real root; if Δ <

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