📚 Solutions of g(x) ≥ f(x) Analytically and Graphically | 解析法与图像法求解 g(x) ≥ f(x)
rt of the IB Mathematics curriculum. Solving them successfully requires shifting between algebraic manipulation and a clear visual understanding of graphs. This article will guide you through both analytical and graphical methods, using a variety of function types to build deep comprehension.
比较两个函数的不等式，例如 g(x) ≥ f(x)，是 IB 数学课程的核心内容之一。要成功求解这类不等式，需要在代数运算与清晰的图形理解之间灵活转换。本文将通过多种函数类型，带你逐步掌握解析法和图像法，建立深刻的理解。
1. Understanding the Problem g(x) ≥ f(x) | 理解问题 g(x) ≥ f(x)
f(x) gives the boundary points, which correspond to intersection points of the two graphs.
不等式 g(x) ≥ f(x) 要求找出所有满足 g 的 y 坐标大于或等于 f 的 y 坐标的 x 值。从几何意义上看，这意味着 g 的图像位于 f 的图像之上或与之重合。等式 g(x) = f(x) 给出了边界点，它们对应于两个图像的交点。
Analytically, we typically rearrange the inequality to h(x) ≥ 0, where h(x) = g(x) – t also be considered, especially if the original functions involve denominators or square roots.
在解析法中，我们通常将不等式变形为 h(x) ≥ 0，其中 h(x) = g(x) – f(x)。这样就将问题转化为研究单个函数 h(x) 的符号，利用临界点可以更容易地进行分析。同时还必须考虑 h 的定义域，尤其当原函数包含分母或平方根时。
2. General Analytical Strategy | 通用解析策略
To solve g(x) ≥ f(x) analytically, follow these steps:
要用解析法求解 g(x) ≥ f(x)，请遵循以下步骤：
Step 1: Form the difference function h(x) = g(x) – f(x). The inequality becomes h(x) ≥ 0.
步骤 1：构造差函数 h(x) = g(x) – f(x)，不等式变为 h(x) ≥ 0。
h(x) = 0 or where h(x) is undefined. These points split the real line into open intervals.
步骤 2：找出所有使得 h(x) = 0 或 h(x) 无定义的临界 x 值。这些点将实数轴分割为若干个开区间。
Step 3: Choose a test value inside each interval and determine the sign of h(x) at that test value. Because h is continuous on each interval (provided no other discontinuities exist), the sign will remain constant throughout the interval.
步骤 3：在每个区间内选取一个测试值，并确定 h(x) 在该测试值处的符号。由于 h 在每个区间上连续（假设没有其他间断点），其符号在整个区间内保持不变。
ction is defined. Write the final solution using inequality notation or interval notation.
步骤 4：列出 h(x) > 0 的区间，并纳入所有使得 h(x) = 0 且函数有定义的端点。用不等式或区间表示法写出最终解。
This sign‑chart method works for polynomial, rational, exponential, and many piecewise functions, though care must be taken with discontinuities and domain restrictions.
这种符号表方法适用于多项式、有理函数、指数函数以及许多分段函数，但需要注意间断点和定义域的限制。
3. Solving Linear Inequalities | 求解线性不等式
Consider g(x) = 2x + 1 and f(x) = x – 3. The inequality 2x + 1 ≥ x – 3 is linear. Using the difference function: h(x) = (2x + 1) – (x – 3) = x + 4 ≥ 0.
考虑 g(x) = 2x + 1 与 f(x) = x – 3。不等式 2x + 1 ≥ x – 3 是线性的。利用差函数：h(x) = (2x + 1) – (x – 3) = x + 4 ≥ 0。
Solving x + 4 ≥ 0 gives x ≥ – gement step instantly yields the answer. Graphically, the line y = 2x + 1 lies above y = x – 3 for all x > –4, and they intersect at x = –4.
解 x + 4 ≥ 0 得到 x ≥ –4。这个简单的例子表明，即使是线性函数，移项步骤也能立刻得出答案。从图像上看，当 x > –4 时，直线 y = 2x + 1 位于 y = x – 3 之上，它们在 x = –4 处相交。
4. Solving Quadratic Inequalities | 求解二次不等式
Let g(x) = x² and f(x) = 4x – 3. The inequality x² ≥ 4x – 3 becomes h(x) = x² – 4x + 3 ≥ 0. Factorising gives (x – 1)(x – 3) ≥ 0. The zeros are x = 1 and x = 3.
设 g(x) = x²，f(x) = 4x – 3。不等式 x² ≥ 4x – 3 变为 h(x) = x² – 4x + 3 ≥ 0。因式分解得 (x – 1)(x – 3) ≥ 0。零点为 x = 1 和 x = 3。
Create a sign table:
建立符号表：
| Interval | Test x | Sign of (x-1)(x-3) |
|---|---|---|
| x < 1 | 0 | (+) · (−) = + |
| 1 < x < 3 | 2 | (+) · (−) = − |
| x > 3 | 4 | (+) · (+) = + |
, the endpoints are included. Thus the solution is x ≤ 1 or x ≥ 3. In interval notation: (–∞, 1] ∪ [3, ∞).
乘积在区间 [1, 3] 之外为正。因为不等式包含等号，端点也应包含在内。因此解为 x ≤ 1 或 x ≥ 3。用区间表示为：(–∞, 1] ∪ [3, ∞)。
ng intervals. Graphically, the parabola y = x² opens upward and intersects the line y = 4x – 3 at x = 1 and x = 3; the parabola is above the line exactly on the outer intervals.
这个二次的例子凸显了因式分解和区间测试的重要性。从图像上看，抛物线 y = x² 开口向上，与直线 y = 4x – 3 在 x = 1 和 x = 3 处相交；抛物线恰好在外部区间位于直线之上。
5. Dealing with Rational Functions | 处理有理函数
For g(x) = 1/(x – 2) and f(x) = 2, the inequality 1/(x – ≥ 2 requires careful handling of the denominator. Bring all terms to one side: h(x) = 1/(x – 2) – 2 ≥ 0. Combine into a single fraction: [1 – 2(x – 2)] / (x – 2) = (5 – 2x) / (x – 2) ≥ 0.
对于 g(x) = 1/(x – 2) 和 f(x) = 2，不等式 1/(x – 2) ≥ 2 需要小心处理分母。将所有项移到一边：h(x) = 1/(x – 2) – 2 ≥ 0。合并为单一分式：[1 – 2(x – 2)] / (x – 2) = (5 – 2x) / (x – 2) ≥ 0。
Critical points come from the numerator (5 – fined). These split the number line into three intervals: x < 2, 2 < x < 2.5, and x > 2.5.
临界点来自分子（5 – 2x = 0 → x = 2.5）和分母（x = 2，在该点表达式无定义）。它们将数轴分为三个区间：x < 2、2 < x < 2.5 以及 x > 2.5。
Test a value in each interval. For x = 0 (x < 2): (5 – 0)/(0 – 2) = 5/(–2) < 0. For x = 2.2 (2 < x < 2.5): (5 – 4.4)/(2.2 – 2) = 0.6/0.2 > 0. For x = 3 (x > 2.5): (5 – 6)/(3 – 2) = –1/1 < 0. The expression is non‑ ecause x = 2 makes the denominator zero it is excluded. Hence the solution is 2 < x ≤ 2.5.
在每个区间内取测试值。当 x = 0（x < 2）：(5 – 0)/(0 – 2) = 5/(–2) < 0。当 x = 2.2（2 < x < 2.5）：(5 – 4.4)/(2.2 – 2) = 0.6/0.2 > 0。当 x = 3（x > 2.5）：(5 – 6)/(3 – 2) = –1/1 < 0。表达式仅在 (2, 2.5] 上非负，由于 x = 2 使分母为零故排除。因此解为 2 < x ≤ 2.5。
When solving rational inequalities, always record all domain restrictions. A common error is to multiply both sides by (x – 2) without considering its sign, which can flip the inequality. The sign‑chart approach avoids this pitfall.
解有理不等式时，务必记录所有的定义域限制。一个常见错误是不考虑 (x – 2) 的符号就直接将两边同乘，这可能会翻转不等号。符号表方法能够避免这个陷阱。
6. Graphical Interpretation | 图像解释
) correspond to intersection points. To identify the solution set, we locate these intersections on the x‑axis and then read off the intervals where one graph is visibly higher than the other.
从图像上看，g(x) ≥ f(x) 表示曲线 y = g(x) 位于曲线 y = f(x) 之上或与之重合。等式 g(x) = f(x) 对应的点就是图像的交点。为了确定解集，我们在 x 轴上找出这些交点，然后读取其中一个图像明显高于另一个图像的区间。
A sketch does not need to be perfectly to scale, but it must correctly show relative positions, intersection points, and key features such as asymptotes or turning points. Technology can quickly generate accurate graphs, but sketching by hand develops intuition.
草图不需要严格按比例绘制，但必须正确展示相对位置、交点以及渐近线或转折点等关键特征。技术工具可以快速生成精确图形，但手绘草图有助于培养直觉。
7. Step‑by‑Step Graphical Method | 分步图像法
y = f(x). If the functions are unfamiliar, build a table of values or use transformations.
步骤 1：在同一坐标系中画出 y = g(x) 和 y = f(x) 的草图。如果函数不熟悉，可以通过数值表或图像变换来构建。
Step 2: Solve g(x) = f(x) to find the x‑oordinates of all intersection points. These are the boundaries of the solution intervals.
步骤 2：解方程 g(x) = f(x)，求出所有交点的 x 坐标。这些就是解区间的边界。
The inequality g(x) ≥ f(x) holds where the graph of g is above or touches the graph of f.
步骤 3：观察在由交点分隔的每个区域内，哪一个图像在上方。在 g 的图像位于 f 图像之上或与之接触的区域，不等式 g(x) ≥ f(x) 成立。
Step 4: Write the answer as a union of intervals. Include the x‑values where the graphs intersect (provided they belong to the domains of both functions).
步骤 4：将答案写成区间的并集。包含两图像相交处的 x 值（前提是它们属于两个函数的定义域）。
This method is particularly powerful when an analytical approach leads to equations that are difficult to solve exactly. The graph reveals the number of solutions and approximate intervals, which can then be refined using a calculator.
当解析法导致难以精确求解的方程时，图像法尤其有效。图像能揭示解的个数和大致区间，之后再利用计算器加以细化。
8. Graphical Example with a Quadratic and a Line | 二次函数与直线的图像示例
Use the earlier example: g(x) = x², f(x) = 4x – 3. Sketch the parabola y = x² and the straight line y = 4x – 3. Their intersection points satisfy x² = 4x – 3, i.e. x² – 4x + 3 = 0, giving x = 1 and x = 3.
使用前面的例子：g(x) = x²，f(x) = 4x – 3。画出抛物线 y = x² 和直线 y = 4x – 3。它们的交点满足 x² = 4x – 3，即 x² – 4x + 3 = 0，解得 x = 1 和 x = 3。
the left of x = 1, the parabola is above the line (for example at x = 0, g(0)=0 and f(0)= –3, so 0 ≥ – parabola is again above. Thus the solution matches the analytical result: x ≤ 1 or x ≥ 3.
在 x = 1 左侧，抛物线位于直线上方（例如 x = 0 时 g(0)=0，f(0)= –3，因此 0 ≥ –3 成立）。在 x = 1 与 x = 3 之间，直线位于抛物线上方（x = 2 时 g(2)=4，f(2)=5，因此 4 ≥ 5 不成立）。在 x = 3 右侧，抛物线再次处于上方。因此解与解析结果一致：x ≤ 1 或 x ≥ 3。
Including intersection points is essential because the inequality is non‑strict. The graphical approach visually confirms why the two outer intervals form the complete solution.
包含交点是必要的，因为不等式含等号。图像法从视觉上确认了为什么两个外部区间构成完整的解集。
9. Handling More Complex Functions | 处理更复杂的函数
logarithms or trigonometric functions, the same principles apply. For instance, to solve eˣ ≥ x + 1 graphically, plot y = eˣ and y = x + 1. The curve y = eˣ it for all other x. Thus the inequality holds for all real x, with equality only at x = 0.
当 g 与 f 包含指数、对数或三角函数时，同样的原则依然适用。例如，用图像法求解 eˣ ≥ x + 1，画出 y = eˣ 和 y = x + 1。曲线 y = eˣ 在 x = 0 处与直线相切，且在其他所有 x 处均位于直线上方。因此，该不等式对所有实数 x 均成立，仅当 x = 0 时取等号。
Analytically, one would consider h(x) = eˣ – x – mediate insight: the exponential curve is convex and lies above its tangent line at x = 0.
在解析法中，我们会考虑 h(x) = eˣ – x – 1，并利用微积分证明它在 x = 0 处取得最小值 0。然而，图像法可以立即提供直观认识：指数曲线是凸的，并且位于其在 x = 0 处的切线之上。
For functions with vertical asymptotes, such as g(x) = ln x and f(x) = 1 – ain: ln x is only defined for x > 0, so any solution set must be restricted to positive x.
对于具有垂直渐近线的函数，例如 g(x) = ln x 和 f(x) = 1 – x，图像能够迅速显示 ln x 在何处位于直线之上。务必检查定义域：ln x 仅在 x > 0 时有定义，因此任何解集都必须限制在正数范围内。
10. Common Mistakes and Pitfalls | 常见错误与陷阱
Mistake 1: Forgetting to check domain restrictions. An inequality like √x ≥ x – the squaring step must be handled with caution to avoid introducing extraneous solutions.
错误 1：忘记检查定义域限制。类似 √x ≥ x – 2 的不等式要求 x ≥ 0，并且平方步骤必须谨慎处理，避免引入增根。
Mistake 2: Multiplying an inequality by an expression whose sign is unknown. Always use the sign‑chart method or bring terms to one side instead of cross‑multiplying wildly.
错误 2：将不等式乘以一个符号未知的表达式。应始终使用符号表方法或称项到一边，而不是随意交叉相乘。
Mistake 3: Misinterpreting ‘ e h(x) = 0 (and are defined) are included. Omitting them can lead to an incomplete answer.
错误 3：误读 “≥”。等号部分意味着 h(x) = 0 且函数有定义的端点需要包含在内。遗漏它们会导致答案不完整。
Mistake 4: Drawing the graphs incorrectly, especially near asymptotes or turning points. A rough sketch that misses a crucial intersection can give the wrong interval. Always verify intersection points algebraically if possible.
错误 4：图像绘制有误，尤其是在渐近线或转折点附近。如果草图漏掉了关键的相交点，可能得出错误的区间。只要可能，总是通过代数方法验证交点。
11. Using Technology: Graphing Calculators | 使用技术：图形计算器
is permitted and can be invaluable. To solve g(x) ≥ f(x), graph both functions and use the ‘intersect’ function to find exact or approximate intersection points. Then use the ‘trace’ or ‘analysis’ features to determine where one graph is above the other.
在 IB 考试中允许使用图形计算器（GDC），且它能发挥巨大作用。要求解 g(x) ≥ f(x)，可画出两个函数的图像，使用“交点”功能找到精确或近似的交点。然后利用“追踪”或“分析”功能确定一个图像在何处位于另一个之上。
When handling functions like sin x and cos x, the periodicity means there may be infinitely many intersection points. The calculator can help you find the pattern and express the solution using periodic intervals.
当处理诸如 sin x 和 cos x 之类的函数时，周期性意味着可能存在无限多个交点。计算器可以帮助你找出规律，并用周期区间表示解。
Do not rely solely on technology; always interpret what the screen shows in the context of the domain and any asymptotes that the calculator might misrepresent (e.g., a vertical asymptote appearing as a ‘false’ vertical line). A solid analytical backup ensures correctness.
不要完全依赖技术；始终结合定义域和渐近线（计算器可能会显示不准确的渐近线，例如垂直渐近线呈现为“伪”垂直线）来解读屏幕上的内容。扎实的解析法备份可以确保正确性。
12. Summary and Key Takeaways | 总结与要点回顾
est intervals, and account for domain restrictions. The method works universally for continuous functions.
解析法中，将 g(x) ≥ f(x) 转化为 h(x) ≥ 0，找出临界点，测试区间，并考虑定义域限制。该方法普遍适用于连续函数。
Graphically, sketch both functions, locate intersection points, and shade the x‑intervals where g sits on or above f. This approach offers a powerful visual check and is essential when exact algebra becomes cumbersome.
图像法中，画出两个函数的草图，标出交点，并标出 g 位于 f 之上或与之重合的 x 区间。这种方法提供了强大的视觉检验，且在精确代数运算变得繁复时不可或缺。
Combining both methods — algebraic precision and graphical intuition — builds robust problem‑solving skills. Whether you face polynomials, rational expressions, or transcendental functions, the framework described here will serve you well throughout the IB course and beyond.
将解析的精确性与图像的直观性结合起来，可以培养扎实的问题解决能力。无论面对多项式、有理式还是超越函数，本文所述的框架都将在整个 IB 课程乃至更远的数学学习中为你提供良好支持。
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