Evaluate Using Function Notation - Algebra-2

1. Fundamental Concepts

Evaluating Using Function Notation: The process of obtaining the corresponding output value (value of the dependent variable) through substitution and calculation when a function expression and a specific input value (value of the independent variable) are given.
Core Symbol Meaning: For a function \(f(x)\), if a given input value is a, then \(f(a)\) represents the result obtained by substituting \(x = a\) into the function expression. It is read as "f of a", meaning "the output of function f when the input is a".

2. Key Concepts

Essence: It is a specific application of the "input-output" correspondence of a function. Through substitution and calculation, it connects the abstract function expression with specific numbers or algebraic expressions.
Types of Inputs:
- Can be specific numerical values (integers, fractions, decimals, etc.);
- Can also be algebraic expressions (such as \(f(a)\), \(f(x + 1)\), etc.).
Consistency with the Definition of a Function: The result of evaluation must satisfy the "single-value correspondence" characteristic of a function, that is, for a definite input, only one output can be obtained.

3. Examples

Easy Level

Example 1: Given the function \(f(x) = 3x + 2\), find \(f(5)\).

Analysis: Substitute \(x = 5\) into the function expression: \(f(5) = 3×5 + 2 = 15 + 2 = 17\).
Conclusion: \(f(5) = 17\).

Medium Level

Example 2: For the function \(g(x) = x^2 - 4x + 3\), find \(g(-2)\) and \(g(b)\).

Analysis:
- When finding \(g(-2)\), substitute \(x = -2\): \(g(-2) = (-2)^2 - 4×(-2) + 3 = 4 + 8 + 3 = 15\);
- When finding \(g(b)\), substitute \(x = b\): \(g(b) = b^2 - 4b + 3\).
Conclusion: \(g(-2) = 15\), \(g(b) = b^2 - 4b + 3\).

Hard Level

Example 3: A courier company's charging standard is: 8 dollars for the first 1kg, and 5 yuan per kg for the part exceeding 1kg (less than 1kg is calculated as 1kg). Let the weight of the item be w kg (\(w \geq 1\)), and the freight \(F(w)\) is a function of w, that is, \(F(w) = 8 + 5×\lceil w - 1 \rceil\) (where \(\lceil x \rceil\) represents the ceiling function). Find the freight for a weight of 3.2kg.

Analysis: The weight \(w = 3.2\) kg, the part exceeding 1kg is \(3.2 - 1 = 2.2\) kg, which is rounded up to 3kg. Substituting into the function: \(F(3.2) = 8 + 5×3 = 8 + 15 = 23\).
Conclusion: \(F(3.2) = 23\).

4. Problem-Solving Techniques

Clarify the function expression and input value: First, determine the known function expression (such as \(f(x) = \text{expression}\)) and the input value to be substituted (such as \(x = a\)).
Substitution and replacement: Replace all independent variable symbols (such as x) in the function expression with the given input value (such as a). Note that when substituting negative numbers or fractions, parentheses should be added to avoid calculation errors (for example, when \(x = -3\), \(x^2\) should be written as \((-3)^2\) instead of \(-3^2\)).
Calculate according to operation rules: Calculate according to the order of algebraic operations (first exponentiation, then multiplication and division, then addition and subtraction; calculate the contents in parentheses first) to ensure the accuracy of each step of the operation.
Handle algebraic expression inputs: When the input is a letter or an algebraic expression, directly retain the letter for substitution without simplification (unless required by the problem). For example, \(f(x + 1)\) only needs to replace x in the expression with \(x + 1\).
Analyze in combination with practical scenarios: In practical problems, first determine the function expression according to the context, clarify the actual meaning and value range of the variable (such as weight, quantity, etc., which are usually non-negative), and then substitute for calculation to ensure that the result conforms to the actual meaning.