高中数学函数单调性与最值综合解题模板!

题目呈现

已知函数 f (x)=x²-2x+3，x∈[0,3]，求函数的单调区间与最值。

Determine the monotonic intervals and the maximum/minimum values of the function f(x)=x²-2x+3 where x∈[0,3].

解题目标

掌握二次函数在闭区间上的单调性判断方法，精准求解函数最值。

Master the method for judging monotonicity of quadratic functions on closed intervals and accurately solve for the maximum and minimum values of functions.

核心原理

二次函数对称轴公式：x = -b/(2a)，结合定义域判断单调性，再比较区间端点与顶点函数值得最值。

Core principle: Use the axis of symmetry formula x = -b/(2a) for quadratic functions, judge monotonicity combined with the domain, and compare function values at endpoints and vertex to obtain extrema.

必备核心知识点

二次函数图像性质、函数单调性定义、闭区间最值求解逻辑。

Key knowledge points: Properties of quadratic function graphs, definition of function monotonicity, logic for solving extrema on closed intervals.

题干已知条件

函数为二次函数，解析式 f (x)=x²-2x+3，定义域为闭区间 [0,3]。

Given conditions: The function is quadratic with 解析式 f (x)=x²-2x+3, and the domain is the closed interval [0,3].

高频易错错解

忽略定义域范围，直接按全体实数判断单调性；仅计算顶点值，未比较区间端点值。

Common mistakes: Ignoring the domain and judging monotonicity over all real numbers; only calculating the vertex value without comparing endpoint values.

纠错避坑依据

二次函数在闭区间上的最值必出现在顶点或区间端点处，需结合定义域限定单调区间。

Basis for error correction: The extrema of a quadratic function on a closed interval must appear at the vertex or interval endpoints; monotonic intervals must be limited by the domain.

规范步骤推导

1. 确定对称轴：x=1；
2. 划分单调区间：[0,1] 上单调递减，[1,3] 上单调递增；
3. 计算函数值：f (1)=2，f (0)=3，f (3)=6；
4. 结论：最小值为 2，最大值为 6。

Standard steps:

1. Determine the axis of symmetry: x=1;
2. Divide monotonic intervals: Decreasing on [0,1], increasing on [1,3];
3. Calculate function values: f(1)=2, f(0)=3, f(3)=6;
4. Conclusion: Minimum value is 2, maximum value is 6.