📚 Solving Inequalities Combined with Functions | IB数学：不等式解法与函数结合
Inequalities are among the most frequently tested topics in IB Mathematics. When combined with functions, they require not only algebraic manipulation but also a deep understanding of graphs, domains, and the behavior of different function families. This article presents a systematic approach to solving inequalities involving functions, with the methods and pitfalls most relevant to IB students.
不等式是IB数学中考查频率最高的主题之一。当不等式与函数结合时，解题不仅需要代数变形，更需要对图像、定义域以及各类函数性质有深刻理解。本文将系统地讲解与函数有关的不等式解法，并聚焦IB考试中最常见的方法与易错点。
1. Fundamental Concepts: f(x) > 0 and f(x) > g(x) | 基本概念：f(x)>0 与 f(x)>g(x)
An inequality involving a single function such as f(x) > 0 asks: for which x-values does the graph of f lie above the x-axis? Similarly, f(x) > g(x) asks where the graph of f is strictly higher than the graph of g.
只含一个函数的不等式，如 f(x) > 0，实际上是在问：x 取何值时，函数 f 的图像位于 x 轴上方？同理，f(x) > g(x) 是在问：x 取何值时，f 的图像严格高于 g 的图像？
Solving these inequalities usually involves three steps:
求解这类不等式通常分为三个步骤：
- Find the critical points where f(x) = 0 or f(x) = g(x); these split the domain into intervals.
- Test each interval using a representative x-value, or examine the sign of the expression in each interval.
- Combine the intervals where the inequality is satisfied, paying attention to strict vs. non-strict conditions.
- 先求出临界点，即 f(x)=0 或 f(x)=g(x) 的解；这些点把定义域分成若干区间。
- 在每个区间内选取代表性 x 值代入验证，或直接分析表达式在各区间的符号。
- 最后合并满足不等式的区间，并注意严格不等式与不严格不等式的区别。
2. Sign Diagrams: Polynomial and Rational Inequalities | 符号图：多项式与有理函数不等式
For polynomial and rational functions, the sign diagram method is powerful. Factor the expression completely, find all zeros and undefined points, then place them on a number line. Between consecutive critical points, the sign of the product or quotient remains constant, provided the function is continuous and has no zeros in between.
对于多项式与有理函数，符号图法非常有效。先将表达式完全因式分解，找出所有零点和无定义点，并标在数轴上。在相邻临界点之间，只要函数连续且无零点，乘积或商的正负号保持不变。
For example, to solve (x-1)(x+2) > 0:
例如，解不等式 (x-1)(x+2) > 0：
(x – 1)(x + 2) > 0
The critical points are x = -2 and x = 1. Testing intervals:
临界点为 x = -2 与 x = 1。分别检验区间：
- x < -2: both factors negative, product positive → solution.
- -2 < x < 1: one negative, one positive → product negative.
- x > 1: both positive → product positive.
- x < -2：两因式均为负，乘积为正 → 可行。
- -2 < x < 1：一负一正，乘积为负。
- x > 1：两因式均为正，乘积为正。
Thus the solution is x < -2 or x > 1. For rational inequalities, remember to exclude values that make the denominator zero, even if the numerator is zero at the same point.
因此解为 x < -2 或 x > 1。对有理函数不等式，务必排除使分母为零的点，即使该点同时使分子为零也不能取。
3. Quadratic Inequalities | 二次不等式
Quadratic inequalities of the form ax² + bx + c > 0 or < 0 are essential. First solve the corresponding quadratic equation ax² + bx + c = 0. The sign of the quadratic depends on the sign of a and the position relative to the roots.
形如 ax² + bx + c > 0 或 < 0 的二次不等式是基础内容。首先解对应的二次方程 ax² + bx + c = 0。二次函数的符号取决于 a 的符号以及 x 相对于两个根的位置。
If a > 0, the parabola opens upward, so the expression is positive outside the roots and negative between the roots. If a < 0, the situation is reversed.
若 a > 0，抛物线开口向上，因此表达式在两根之外为正、两根之间为负；若 a < 0，情况正好相反。
Consider x² – 5x + 6 ≤ 0. Factoring gives (x – 2)(x – 3) ≤ 0. The roots are 2 and 3. Since the leading coefficient is positive, the expression is ≤ 0 between the roots, so the solution is 2 ≤ x ≤ 3.
考虑 x² – 5x + 6 ≤ 0。因式分解得 (x – 2)(x – 3) ≤ 0。两根为 2 和 3。由于二次项系数为正，表达式在两根之间 ≤ 0，因此解为 2 ≤ x ≤ 3。
When the quadratic has no real roots, its sign is always the same as a. When it has a double root, the expression touches zero at one point but does not change sign.
当二次方程无实根时，二次函数的符号恒与 a 相同；当方程有二重根时，函数在该点恰好为零但符号不改变。
4. Absolute Value Inequalities | 绝对值不等式
Absolute value functions create piecewise behavior. The key identity is |u| > c ⇔ u > c or u < -c, while |u| < c ⇔ -c < u < c when c > 0. But when u is itself a function, one must also consider its zeros and domain.
绝对值函数具有分段性质。关键恒等式为：当 c > 0 时，|u| > c ⇔ u > c 或 u < -c，而 |u| < c ⇔ -c < u < c。但当 u 本身是函数时，还需考虑其零点和定义域。
For an inequality involving |f(x)| < g(x), we can rewrite it as -g(x) < f(x) < g(x), provided g(x) > 0 for all relevant x. If g(x) can be negative, the inequality may have no solution, or a more careful analysis is required.
对于含 |f(x)| < g(x) 的不等式，可改写成 -g(x) < f(x) < g(x)，前提是相关区间内 g(x) > 0。若 g(x) 可能为负，则不等式可能无解，或需要更仔细的分析。
For example, solve |2x – 1| < 3. This gives -3 < 2x - 1 < 3. Adding 1 and dividing by 2 yields -1 < x < 2.
例如，解 |2x – 1| < 3。可得 -3 < 2x - 1 < 3，加 1 再除以 2，得 -1 < x < 2。
A common IB challenge is solving |f(x)| > |g(x)|. Squaring both sides is valid because both sides are non-negative, leading to f(x)² > g(x)². This often simplifies to a polynomial inequality.
一个常见的IB考点是解 |f(x)| > |g(x)|。两边都是非负的，因此可以安全地平方，得到 f(x)² > g(x)²，这通常能化简为多项式不等式。
5. Rational Inequalities: Avoiding Multiplication Pitfalls | 有理不等式：避免乘除陷阱
A major source of errors is multiplying both sides of a rational inequality by a quantity that may be negative. For instance, to solve 1/x > 2, students often multiply by x without knowing its sign. Instead, move all terms to one side and combine into a single fraction.
有理不等式最大的错误来源是“两边同乘”一个符号不确定的量。例如解 1/x > 2 时，学生常在不知道 x 正负的情况下直接乘 x。正确做法是把所有项移到一边，合并为一个分式。
Rewrite 1/x > 2 as 1/x – 2 > 0, i.e., (1 – 2x)/x > 0. The critical points are x = 0 and x = 1/2. A sign diagram shows the solution is 0 < x < 1/2.
将 1/x > 2 改写为 1/x – 2 > 0，即 (1 – 2x)/x > 0。临界点为 x = 0 与 x = 1/2。符号图显示解为 0 < x < 1/2。
In general, for any rational inequality, never multiply by a denominator that could be zero or negative. Instead, bring all terms to one side, factor numerator and denominator, and analyze signs with a diagram.
一般来说，对于任何有理不等式，绝不要乘以可能为零或为负的分母。应把所有项移到一边，对分子分母分别因式分解，然后用符号图分析正负。
6. Graphical Interpretation and Intersection Points | 图像解释与交点
Graphical methods are especially useful for non-algebraic functions or when the inequality combines different function families. To solve f(x) > g(x), draw both graphs and locate their intersection points. The solution is the x-interval where the graph of f lies above the graph of g.
对于无法直接代数求解或涉及不同函数族的不等式，图像法尤其有用。要解 f(x) > g(x)，画出两个函数图像并求交点。满足条件的 x 区间就是 f 的图像位于 g 的图像上方的部分。
With a GDC (Graphic Display Calculator), IB students can quickly find intersections numerically and then read the required intervals. However, exact algebraic answers are often required, so use the graph as a check rather than the final proof.
使用图形计算器（GDC），IB学生可以快速求得交点数值，并直接读出所需区间。但考试常常要求精确代数答案，因此图像应作为验证工具，而非最终证明。
When solving f(x) ≥ 0, the x-axis itself is the reference line. The zeros of f are the intersections with the x-axis, and the sign of f determines whether the graph is above or below the axis.
解 f(x) ≥ 0 时，x 轴本身是参考线。f 的零点就是与 x 轴的交点，f 的符号决定了图像在轴上方还是下方。
7. Transformations: Squaring, Inverting and Exponential Mapping | 变换：平方、取倒数与指数映射
Some inequalities require applying a function to both sides. This is valid only if the function is monotonic on the relevant interval. For example, squaring both sides is valid if both sides are non-negative. If both sides are negative, squaring reverses the inequality because the square function is decreasing on (-∞, 0).
有些不等式需要对两边同时施加某个函数。这仅在函数于相关区间内单调时才有效。例如，两边平方要求两边均为非负；若两边为负，由于平方函数在 (-∞, 0) 上递减，平方后不等式方向会反转。
Consider solving √(x+1) < 2. Since the square root is increasing, we can square both sides: x + 1 < 4, giving x < 3. But we must also require x + 1 ≥ 0, hence x ≥ -1. The final solution is -1 ≤ x < 3.
例如解 √(x+1) < 2。因为平方根函数单调递增，两边平方得 x + 1 < 4，即 x < 3。但还须要求 x + 1 ≥ 0，所以 x ≥ -1。最终解为 -1 ≤ x < 3。
Similarly, the natural logarithm ln x is increasing, so f(x) < g(x) implies ln(f(x)) < ln(g(x)) as long as f(x) > 0 and g(x) > 0. The exponential function eˣ is also increasing, preserving direction.
类似地，自然对数 ln x 单调递增，因此当 f(x) > 0 且 g(x) > 0 时，f(x) < g(x) 可推出 ln(f(x)) < ln(g(x))。指数函数 eˣ 也是增函数，不改变不等式方向。
Taking reciprocals is a more delicate operation: 1/f(x) < 1/g(x) does not simply follow. The reciprocal function is decreasing on each interval separated by zero, but the inequality direction flips depending on the signs of f and g.
取倒数则更需谨慎：1/f(x) < 1/g(x) 不能简单套用普通规则。倒数函数在每个由零点分隔的区间上单调递减，但方向翻转取决于 f 和 g 的符号。
8. Compound and Systems of Inequalities | 复合不等式与不等式组
Sometimes an inequality appears as a double inequality, such as 1 < f(x) < 5. This is a conjunction of two inequalities: f(x) > 1 and f(x) < 5. The solution is the intersection of the two solution sets.
有时不等式以双重不等式形式出现，如 1 < f(x) < 5。这是两个不等式的合取：f(x) > 1 与 f(x) < 5。最终解为两个解集的交集。
For systems of inequalities like f(x) > 0 and g(x) > 0, solve each separately and then take the intersection of the resulting intervals. If the system uses “or”, combine the sets as a union.
对于不等式组，例如 f(x) > 0 且 g(x) > 0，先分别解每个不等式，再对区间取交集；如果是“或”的关系，则取并集。
In IB, compound inequalities often appear implicitly in functions problems, such as determining the domain of h(x) = √(f(x)) where f(x) ≥ 0, or h(x) = ln(f(x)) where f(x) > 0.
在IB中，复合不等式常以隐含形式出现在函数问题里，比如求 h(x) = √(f(x)) 的定义域时需解 f(x) ≥ 0，求 h(x) = ln(f(x)) 的定义域时需解 f(x) > 0。
9. Exponential and Logarithmic Inequalities | 指数与对数不等式
For exponential inequalities such as aˣ > b, take logarithms of both sides. If the base a > 1, logₐ is increasing, so direction is preserved. If 0 < a < 1, the logarithm is decreasing, so the inequality flips.
对于指数不等式，如 aˣ > b，两边取对数。若底数 a > 1，logₐ 单调递增，不等式方向不变；若 0 < a < 1，对数是减函数，方向需要反转。
Example: 2ˣ > 8. Since 8 = 2³, we get x > 3 directly. For a more complex case, 3ˣ < 5 transforms to x < log₃ 5.
例：2ˣ > 8。因为 8 = 2³，直接得到 x > 3。对于更复杂的情况，3ˣ < 5 可转化为 x < log₃ 5。
Logarithmic inequalities require checking the domain carefully: log₂(x – 1) > 3 means x – 1 > 8, so x > 9. The original argument x – 1 must be positive, which is already satisfied by the solution.
对数不等式需要仔细检查定义域：log₂(x – 1) > 3 意味着 x – 1 > 8，所以 x > 9。原函数的真数 x – 1 必须为正，而该解本身已满足这一条件。
When variables appear inside multiple logarithms, combine them using logarithm rules, but be aware that combining logₐ A + logₐ B into logₐ(AB) creates a new expression whose domain may be larger. Always check the original domain.
当多个对数中出现变量时，可用对数运算法则合并，但要注意 logₐ A + logₐ B 合并为 logₐ(AB) 后，定义域可能扩大。一定要以原表达式的定义域为准。
10. Inequalities with Trigonometric Functions | 含三角函数的不等式
Trigonometric inequalities like sin x > 0.5 require knowledge of the unit circle and periodicity. On a given interval, locate the reference angles and then express the solution in terms of periodic intervals.
像 sin x > 0.5 这样的三角不等式需要借助单位圆和周期性。在给定区间内，先找出参考角，然后用周期区间表达解。
For example, sin x > 1/2 on [0, 2π) gives x ∈ (π/6, 5π/6). Since sine has period 2π, the general solution is x ∈ (π/6 + 2πk, 5π/6 + 2πk) for k ∈ ℤ.
例如，在 [0, 2π) 上解 sin x > 1/2，得到 x ∈ (π/6, 5π/6)。由于正弦函数的周期为 2π，通解为 x ∈ (π/6 + 2πk, 5π/6 + 2πk)，其中 k ∈ ℤ。
For inequalities involving both a trigonometric function and another function, such as cos x > x², graphical methods are usually the only feasible approach. IB questions often restrict the domain to assist with exact or GDC-based solutions.
当不等式同时包含三角函数和其他函数时，如 cos x > x²，图像法通常是唯一可行的方法。IB题目通常会限定区间，以帮助进行精确求解或使用图形计算器求解。
11. Common Pitfalls in IB Examinations | IB考试常见易错点
One frequent mistake is forgetting to exclude points where the denominator is zero or where a square root argument is negative. Another is reversing an inequality when multiplying by a negative number, especially inside absolute-value or rational expressions.
一个常见错误是忘记排除分母为零或根号内为负的点。另一个错误是乘以负数时忘记改变不等号方向，尤其是在绝对值或有理表达式中。
When using the sign diagram, be sure to mark “undefined” points with a vertical asymptote; the sign may change across an asymptote even when neither side touches zero. For rational functions, a zero of the denominator breaks the sign pattern.
使用符号图时，务必用垂直渐近线标记“无定义”点；跨越渐近线时符号可能改变，尽管两侧都不为零。对于有理函数，分母的零点会使符号模式中断。
Another common error is to divide by a variable expression without considering whether it could be zero. For example, dividing both sides of x(x-2) > 0 by x is incorrect because x may be negative or zero; factor and use a sign diagram instead.
另一个常见错误是除以含变量的表达式而不考虑其可能为零。例如，解 x(x-2) > 0 时两边同除以 x 是错误的，因为 x 可能为负或零；应当因式分解并用符号图求解。
Finally, always compare with the original inequality: after squaring, take reciprocals, or applying logarithms, verify that all solutions satisfy the original domain and inequality condition.
最后，务必与原不等式对照验证：经过平方、取倒数或对数运算后，要检查所有解是否满足原定义域和不等式条件。
12. Practice and Summary | 练习与总结
To master inequalities with functions, practice the following types:
要掌握函数不等式，建议练习以下类型：
- Polynomial inequalities with repeated roots: (x-1)²(x+3) > 0
- Rational inequalities with asymptotes: (x²-1)/(x-2) ≤ 0
- Absolute value inside a function: |x² – 4| > 3
- Inequalities involving domains: √(x+2) ≤ x
- Mixed functions: eˣ < 2x + 1 (solve graphically)
- 含重根的多项式不等式：(x-1)²(x+3) > 0
- 含渐近线的有理不等式：(x²-1)/(x-2) ≤ 0
- 函数内嵌绝对值：|x² – 4| > 3
- 涉及定义域的不等式：√(x+2) ≤ x
- 混合函数不等式：eˣ < 2x + 1（图像法求解）
In summary, solving inequalities combined with functions requires a clear strategy: identify the function family, determine critical points, examine signs or graphs, and always respect the domain. With careful practice and attention to detail, these problems become predictable and rewarding.
总而言之，解决函数不等式需要清晰的策略：识别函数类型，确定临界点，检查符号或图像，并始终尊重定义域。只要认真练习并注意细节，这类问题就会变得有章可循、得心应手。
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