函数的性质与图像是 AIME 中的核心概念，涉及函数的单调性、奇偶性、周期性、对称性等重要内容。
Properties and graphs of functions are core concepts in AIME, involving monotonicity, parity, periodicity, symmetry, and other important properties of functions.
函数的图像变换：
- 平移变换：y = f(x + h) + k（水平平移 h，垂直平移 k）
- 伸缩变换：y = a·f(b·x)（垂直伸缩 a，水平伸缩 1/b）
- 对称变换：y = -f(x)（关于 x 轴对称），y = f(-x)（关于 y 轴对称）
- 反射变换：y = f(|x|)（保留 y 轴右侧图像，左侧对称）
例题 1 Example 1
这是一个线性递推关系，我们可以计算前几项：
f(1) = 2
f(2) = 3
f(3) = 2f(2) - f(1) = 2×3 - 2 = 4
f(4) = 2f(3) - f(2) = 2×4 - 3 = 5
f(5) = 2f(4) - f(3) = 2×5 - 4 = 6
观察规律，f(n) = n + 1，所以 f(5) = 6。
This is a linear recurrence relation. Calculating the first few terms: f(1) = 2, f(2) = 3, f(3) = 4, f(4) = 5, f(5) = 6. The pattern is f(n) = n + 1, so f(5) = 6.
方程的解法是 AIME 中的重要内容，涉及代数方程、超越方程、方程组等多种类型的方程求解。
Solution of equations is an important content in AIME, involving solving algebraic equations, transcendental equations, systems of equations, and other types of equations.
方程组的解法：
- 代入法：将一个方程代入另一个方程
- 消元法：通过加减消去一个变量
- 矩阵法：利用矩阵的行列式和逆矩阵
- 对称性：利用变量的对称性
例题 2 Example 2
令 t = 2^x，则 4^x = (2^x)^2 = t²，原方程变为：
t + t² = 6
整理得：t² + t - 6 = 0
因式分解：(t + 3)(t - 2) = 0
解得 t = -3 或 t = 2
由于 t = 2^x > 0，所以 t = 2，即 2^x = 2，解得 x = 1。
Let t = 2^x, then 4^x = t². The equation becomes t + t² = 6 → t² + t - 6 = 0 → (t + 3)(t - 2) = 0. Since t = 2^x > 0, t = 2 → 2^x = 2 → x = 1.
函数的应用是 AIME 中的重要内容，涉及函数在几何、物理、经济等领域的应用，以及函数的最大值、最小值问题。
Applications of functions are an important content in AIME, involving the use of functions in geometry, physics, economics, and other fields, as well as problems of maximum and minimum values of functions.
函数的应用场景：
- 几何中的最值问题
- 物理中的运动问题
- 经济中的优化问题
- 组合数学中的计数问题
例题 3 Example 3
求导数 f'(x) = 3x² - 6x + 2
令 f'(x) = 0，解得 x = [6 ± √(36 - 24)]/6 = [6 ± √12]/6 = [6 ± 2√3]/6 = 1 ± (√3)/3
在区间 [0, 3] 内，两个临界点都存在：
x₁ = 1 - (√3)/3 ≈ 0.4226
x₂ = 1 + (√3)/3 ≈ 1.5774
计算函数在临界点和区间端点的值：
f(0) = 0
f(x₁) = (1 - (√3)/3)³ - 3(1 - (√3)/3)² + 2(1 - (√3)/3) ≈ 0.384
f(x₂) = (1 + (√3)/3)³ - 3(1 + (√3)/3)² + 2(1 + (√3)/3) ≈ -0.384
f(3) = 27 - 27 + 6 = 6
所以，函数在区间 [0, 3] 上的最大值为 6。
Find the derivative f'(x) = 3x² - 6x + 2. Set f'(x) = 0, solve for x = 1 ± (√3)/3. Evaluate f(x) at critical points and endpoints: f(0) = 0, f(1 - (√3)/3) ≈ 0.384, f(1 + (√3)/3) ≈ -0.384, f(3) = 6. The maximum value is 6.