📚 Solving g(x) ≥ f(x) Analytically and Graphically | 解析与图解求解 g(x)≥f(x)
In IB Mathematics, inequalities of the form g(x) ≥ f(x) appear frequently in both Analysis and Approaches (AA) and Applications and Interpretation (AI) courses. Understanding how to solve these inequalities both graphically and analytically is essential for examinations, as well as for developing a deeper intuition about functions and their behaviour.
在IB数学中，形如 g(x) ≥ f(x) 的不等式在数学分析与方法（AA）以及应用与解释（AI）课程中都频繁出现。掌握图解法和解析法求解这类不等式，对于考试至关重要，同时也能帮助你更深入地理解函数的性质与变化规律。
1. Understanding the Meaning of g(x) ≥ f(x) | 理解 g(x) ≥ f(x) 的含义
Before attempting to solve any inequality, it is crucial to understand what the statement actually means. The expression g(x) ≥ f(x) asks us to find all values of x for which the value of the function g is greater than or equal to the value of the function f. In other words, we are looking for the set of x-coordinates where the graph of g lies on or above the graph of f.
在尝试求解任何不等式之前，首先要理解这个表达式的真正含义。表达式 g(x) ≥ f(x) 要求我们找出所有使函数 g 的值大于或等于函数 f 的值的 x 坐标。换句话说，我们寻找的是图像 g 位于图像 f 上方或与之重合的所有 x 值。
It is important to note that this is not the same as solving g(x) = f(x). Equality only gives us the boundary points where the two functions intersect. The inequality includes these points as well as the entire region between and around them.
需要注意的是，这并不等同于求解 g(x) = f(x)。等式只给出两个函数相交的边界点，而不等式则包括这些交点以及它们之间和周围的整个区域。
2. The Graphical Approach | 图解法
The graphical method is often the most intuitive way to solve g(x) ≥ f(x). To use this method, follow these steps:
图解法通常是求解 g(x) ≥ f(x) 最直观的方法。使用这一方法时，请遵循以下步骤：
- Step 1: Sketch the graphs of y = f(x) and y = g(x) on the same set of axes.
- Step 2: Identify the x-coordinates of all intersection points by solving f(x) = g(x).
- Step 3: Determine which function has larger y-values on each interval between consecutive intersection points.
- Step 4: Write the solution as the union of intervals where g(x) is above or equal to f(x).
步骤1：在同一坐标系中绘制 y = f(x) 和 y = g(x) 的图像。
步骤2：通过解 f(x) = g(x) 找出所有交点的 x 坐标。
步骤3：确定在相邻交点之间的每个区间上，哪个函数的 y 值更大。
步骤4：将解写为 g(x) 位于 f(x) 上方或与之重合的区间的并集。
For example, consider f(x) = x² and g(x) = 2x + 3. The graphs intersect when x² = 2x + 3, which gives x² − 2x − 3 = 0, so x = 3 or x = −1. Testing a point in each interval shows that g(x) ≥ f(x) when −1 ≤ x ≤ 3.
例如，考虑 f(x) = x² 和 g(x) = 2x + 3。两图像在 x² = 2x + 3 时相交，即 x² − 2x − 3 = 0，解得 x = 3 或 x = −1。在每个区间上测试一个点可以发现，当 −1 ≤ x ≤ 3 时 g(x) ≥ f(x)。
3. The Analytical Approach: Rearranging the Inequality | 解析法：不等式变形
The analytical method involves algebraic manipulation. One common strategy is to rearrange the inequality into the form h(x) ≥ 0, where h(x) = g(x) − f(x). Then, the problem reduces to finding where the function h is non-negative.
解析法涉及代数变形。一种常用策略是将不等式变形为 h(x) ≥ 0 的形式，其中 h(x) = g(x) − f(x)。这样，问题就简化为寻找函数 h 为非负的区域。
This approach is particularly useful for polynomials and rational functions, where sign analysis can be performed systematically. For example, to solve x³ > 4x, rewrite as x³ − 4x > 0, factorise as x(x − 2)(x + 2) > 0, and then use a sign diagram to determine the intervals.
这一方法对多项式函数和有理函数尤为有用，因为可以系统地进行符号分析。例如，求解 x³ > 4x 时，将其改写为 x³ − 4x > 0，分解因式为 x(x − 2)(x + 2) > 0，然后使用符号图表确定区间。
h(x) = g(x) − f(x) ≥ 0
The sign of h(x) can only change at points where h(x) = 0 or where h(x) is undefined. These points divide the real line into intervals, and we test a single value in each interval to determine the sign throughout.
h(x) 的符号只能在 h(x) = 0 或 h(x) 无定义的点处发生变化。这些点将实数轴划分为若干区间，我们在每个区间内取一个测试值来判断整个区间的符号。
4. Linear Inequalities | 线性不等式
When both f and g are linear functions, the inequality reduces to a simple linear inequality. For instance, solve 3x − 2 ≥ x + 4. Subtracting x from both sides gives 2x − 2 ≥ 4, and adding 2 gives 2x ≥ 6, so x ≥ 3.
当 f 和 g 都是线性函数时，不等式简化为简单的一次不等式。例如，求解 3x − 2 ≥ x + 4。两边减去 x 得到 2x − 2 ≥ 4，再加 2 得到 2x ≥ 6，因此 x ≥ 3。
Graphically, this corresponds to the region where the line y = 3x − 2 lies above the line y = x + 4. The intersection point is found at x = 3, and since the slope of the first line is greater, it stays above to the right.
图解上，这对应于直线 y = 3x − 2 位于直线 y = x + 4 上方的区域。交点位于 x = 3，由于第一条直线的斜率更大，它在右侧保持在上方。
5. Quadratic Inequalities | 二次不等式
Quadratic inequalities are extremely common in IB examinations. Consider g(x) = x² − 5x + 6 and f(x) = 0. We wish to solve x² − 5x + 6 ≥ 0. Factorising gives (x − 2)(x − 3) ≥ 0. The critical values are x = 2 and x = 3. Testing intervals yields x ≤ 2 or x ≥ 3.
二次不等式在IB考试中极其常见。考虑 g(x) = x² − 5x + 6 和 f(x) = 0。我们需要求解 x² − 5x + 6 ≥ 0。分解因式得到 (x − 2)(x − 3) ≥ 0。关键值为 x = 2 和 x = 3。测试各区间可得 x ≤ 2 或 x ≥ 3。
A sign diagram is particularly helpful here. Mark the critical values on a number line, then determine the sign of the quadratic in each region. Since the leading coefficient is positive, the quadratic opens upward, so it is positive outside the roots and negative between them.
符号图表在此处特别有用。在数轴上标出关键值，然后判断二次函数在每个区域的符号。由于首项系数为正，抛物线开口向上，因此在两个根之外为正，在根之间为负。
ax² + bx + c ≥ 0 ⇔ x ≤ α 或 x ≥ β（当 a > 0）
If the quadratic has no real roots, then its sign never changes. In this case, the inequality is either always true or never true, depending on the sign of the leading coefficient and the direction of the inequality.
如果二次函数没有实数根，那么它的符号不会改变。在这种情况下，不等式要么恒成立，要么恒不成立，具体取决于首项系数的符号和不等式的方向。
6. Using the Discriminant | 利用判别式
The discriminant Δ = b² − 4ac plays a key role in determining the number of intersection points between a quadratic function and a horizontal line. When solving g(x) ≥ f(x) where g is quadratic and f is linear, the discriminant of the resulting quadratic equation tells us how many intersection points exist.
判别式 Δ = b² − 4ac 在确定二次函数与水平线的交点数方面起着关键作用。当求解 g(x) ≥ f(x) 且 g 为二次函数、f 为一次函数时，所得二次方程的判别式告诉我们交点的个数。
For example, to find whether g(x) = x² − 3x + 2 is always above f(x) = −1, solve x² − 3x + 2 = −1, which becomes x² − 3x + 3 = 0. The discriminant is 9 − 12 = −3 < 0, so there are no real intersections. Since the quadratic opens upward and its vertex is above the line, the inequality holds for all real x.
例如，要判断 g(x) = x² − 3x + 2 是否恒在 f(x) = −1 上方，解 x² − 3x + 2 = −1，得到 x² − 3x + 3 = 0。判别式为 9 − 12 = −3 < 0，因此没有实数交点。由于抛物线开口向上且顶点在直线上方，该不等式对所有实数 x 都成立。
Δ > 0: 两个交点 | Δ = 0: 一个交点 | Δ < 0: 无交点
7. Rational Inequalities | 有理不等式
For rational functions of the form g(x) = P(x)/Q(x), solving g(x) ≥ f(x) requires extra care because the denominator can change sign. The critical values include both the zeros of the numerator and the zeros of the denominator.
对于形如 g(x) = P(x)/Q(x) 的有理函数，求解 g(x) ≥ f(x) 需要格外小心，因为分母可能改变符号。关键值既包括分子的零点，也包括分母的零点。
Consider solving 1/x ≥ x. Rewrite as 1/x − x ≥ 0, giving (1 − x²)/x ≥ 0. Factorise as (1 − x)(1 + x)/x ≥ 0. The critical values are x = −1, x = 0, and x = 1. A sign diagram over the intervals (−∞, −1), (−1, 0), (0, 1), and (1, ∞) reveals that the inequality holds for x ≤ −1 and 0 < x ≤ 1.
考虑求解 1/x ≥ x。将其改写为 1/x − x ≥ 0，得到 (1 − x²)/x ≥ 0。分解为 (1 − x)(1 + x)/x ≥ 0。关键值为 x = −1、x = 0 和 x = 1。在区间 (−∞, −1)、(−1, 0)、(0, 1) 和 (1, ∞) 上制作符号图表可以发现，不等式在 x ≤ −1 和 0 < x ≤ 1 时成立。
Note that x = 0 is excluded from the solution because the original expression is undefined there. This is a common error: forgetting to exclude values that make the denominator zero.
注意 x = 0 被排除在解集之外，因为原表达式在此处无定义。这是一个常见错误：忘记排除使分母为零的值。
8. Absolute Value Inequalities | 绝对值不等式
When functions involve absolute values, the analytical approach often requires splitting the problem into cases. For example, solve |x − 1| ≥ x − 3. Since |x − 1| equals x − 1 when x ≥ 1, and equals 1 − x when x < 1, we consider two separate cases.
当函数涉及绝对值时，解析法通常需要分情况讨论。例如，求解 |x − 1| ≥ x − 3。由于当 x ≥ 1 时 |x − 1| = x − 1，当 x < 1 时 |x − 1| = 1 − x，我们分别考虑两种情况。
For x ≥ 1, the inequality becomes x − 1 ≥ x − 3, which simplifies to −1 ≥ −3, always true. For x < 1, it becomes 1 − x ≥ x − 3, giving 4 ≥ 2x, so x ≤ 2. Combining both cases, the solution is all real x, because the condition x < 1 with x ≤ 2 is simply x < 1, and together with x ≥ 1 gives the entire real line.
当 x ≥ 1 时，不等式变为 x − 1 ≥ x − 3，化简为 −1 ≥ −3，恒成立。当 x < 1 时，变为 1 − x ≥ x − 3，即 4 ≥ 2x，所以 x ≤ 2。综合两种情况，解为所有实数 x，因为 x < 1 且 x ≤ 2 的条件就是 x < 1，与 x ≥ 1 合并后覆盖整个实数轴。
Graphically, this is verified by noting that the V-shaped graph of |x − 1| lies everywhere above the line x − 3, except it touches at no point, so the inequality holds everywhere.
图解上，可以看到 |x − 1| 的 V 形图像处处位于直线 x − 3 上方，且没有交点，因此不等式处处成立。
9. Exponential and Logarithmic Inequalities | 指数与对数不等式
For exponential functions, the base determines the direction of the inequality after taking logarithms. If the base is greater than 1, the logarithm is increasing, so the inequality direction is preserved. If the base is between 0 and 1, the logarithm is decreasing, so the direction is reversed.
对于指数函数，底数决定了取对数后不等式的方向。如果底数大于 1，对数函数是递增的，因此不等式方向保持不变。如果底数在 0 和 1 之间，对数函数是递减的，因此不等式方向需要反转。
For example, solve 2ˣ ≥ 8. Since 8 = 2³, we have 2ˣ ≥ 2³. Because the base 2 > 1, x ≥ 3. For logarithms, solve log₂(x) ≥ 3. This gives x ≥ 2³ = 8. However, we must also remember the domain restriction x > 0, which is automatically satisfied here.
例如，求解 2ˣ ≥ 8。由于 8 = 2³，即 2ˣ ≥ 2³。因为底数 2 > 1，所以 x ≥ 3。对于对数，求解 log₂(x) ≥ 3。这给出 x ≥ 2³ = 8。但别忘了定义域限制 x > 0，这里已自动满足。
When mixing exponential functions with linear functions, analytical solutions are usually impossible in closed form. In such cases, the graphical method or numerical methods using a graphing display calculator (GDC) are required.
当指数函数与线性函数混合时，通常无法以封闭形式获得解析解。在这种情况下，需要使用图解法或图形计算器（GDC）进行数值求解。
10. Working with Parameterised Functions | 含参函数的处理
In IB examinations, you may be asked to find the range of a parameter for which g(x) ≥ f(x) holds for all x or for a specific interval. This type of problem combines inequality solving with function analysis.
在IB考试中，你可能会被要求找出参数的取值范围，使得 g(x) ≥ f(x) 对所有 x 或在特定区间上恒成立。这类问题将不等式求解与函数分析相结合。
For instance, suppose g(x) = x² + kx + 1 and f(x) = 2x − 3. We want the inequality x² + kx + 1 ≥ 2x − 3 to hold for all real x. Rearranging gives x² + (k − 2)x + 4 ≥ 0. For this quadratic to be non-negative for all x, it must have no real roots and open upward. This requires the discriminant to be less than or equal to zero:
例如，设 g(x) = x² + kx + 1，f(x) = 2x − 3。我们希望不等式 x² + kx + 1 ≥ 2x − 3 对所有实数 x 恒成立。变形得到 x² + (k − 2)x + 4 ≥ 0。要使该二次式对所有 x 非负，它必须没有实数根且开口向上。这要求判别式小于或等于零：
(k − 2)² − 16 ≤ 0 ⇔ (k − 2)² ≤ 16 ⇔ −2 ≤ k ≤ 6
This result means the parabola touches the x-axis at the endpoints of this parameter range and lies above it in between.
这个结果表明，当参数取区间端点值时抛物线恰与 x 轴相切，在端点之间则位于 x 轴上方。
11. Sign Diagrams | 符号图表
Sign diagrams are a powerful visual tool for solving inequalities. To construct one, first identify all critical values where h(x) = 0 or h(x) is undefined. Then, draw a horizontal line representing the x-axis and mark these critical points. Finally, test one value in each interval and record whether h(x) is positive, negative, or zero.
符号图表是求解不等式的强大视觉工具。构造步骤如下：首先找出所有 h(x) = 0 或 h(x) 无定义的关键值；然后画一条水平线代表 x 轴并标出这些关键点；最后在每个区间内取一个测试值并记录 h(x) 是正、是负还是零。
A simple example: solve x(x − 1)(x + 2) > 0. The critical values are x = 0, x = 1, and x = −2. The sign changes at each simple root. The sign diagram shows positive values on (−2, 0) ∪ (1, ∞), so that is the solution set.
一个简单的例子：求解 x(x − 1)(x + 2) > 0。关键值为 x = 0、x = 1 和 x = −2。在每个单根处符号发生改变。符号图表显示在 (−2, 0) ∪ (1, ∞) 上取正值，因此这就是解集。
For repeated roots, the sign may not change. For example, (x − 1)² ≥ 0 holds for all real x because a square is always non-negative, even though the expression equals zero at x = 1.
对于重根，符号可能不发生变化。例如，(x − 1)² ≥ 0 对所有实数 x 都成立，因为平方总为非负，尽管该表达式在 x = 1 处等于零。
12. GDC Techniques | 图形计算器技巧
In both AA and AI courses, the graphing display calculator (GDC) is an essential tool. To solve g(x) ≥ f(x) on a GDC:
在AA和AI课程中，图形计算器（GDC）都是必不可少的工具。在GDC上求解 g(x) ≥ f(x) 的方法是：
- Method 1: Graph both functions and use the intersection feature to find points of intersection. Then visually determine the solution intervals.
- Method 2: Define h(x) = g(x) − f(x) and graph only h(x). Find where h(x) ≥ 0 by using the zero or root feature.
- Method 3: Use the inequality graphing mode if available on your GDC model.
方法1：绘制两个函数的图像，使用交点功能找出交点，然后通过观察确定解区间。
方法2：定义 h(x) = g(x) − f(x) 并只绘制 h(x) 的图像。使用零点或求根功能找出 h(x) ≥ 0 的区域。
方法3：如果你的GDC型号支持，可以使用不等式绘图模式。
Remember to set appropriate viewing windows and use smaller step sizes near intersection points to avoid missing any crucial features. Always confirm your GDC results with analytical reasoning whenever possible.
请记住设置合适的视窗范围，并在交点附近使用较小的步长，以免遗漏关键特征。在可能的情况下，务必用解析推理来验证GDC的结果。
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