Case Analysis Flashcards

Analyze by cases: for the equation |x - 3| = 2x, which values of x are valid solutions?

Answer: x = 1 only

Case 1: x - 3 = 2x gives x = -3, but 2(-3) = -6 < 0 so invalid; Case 2: -(x-3) = 2x gives x = 1, check: |1-3| = 2 = 2(1) ✓.

A piecewise function f(x) = -x if x 0. This function is best described as:

Answer: The absolute value function |x|

The definition matches |x| exactly: negative inputs become positive, zero stays zero, positive inputs remain unchanged.

Solve by case analysis: if x² > 9, what is the solution set?

Answer: x 3

x² > 9 means |x| > 3, which splits into x > 3 or x < -3.

A store charges $5 per item for 1–10 items and $4 per item for more than 10 items. Using case analysis, what is the cost for exactly 10 items?

Answer: $50

Exactly 10 items falls in the first pricing tier (1–10 items), so cost = 10 × $5 = $50.

For f(x) = |x - 2| + |x + 1|, what is the minimum value of f(x)?

Answer: 3

For -1 ≤ x ≤ 2, f(x) = (2 - x) + (x + 1) = 3 (constant); outside this interval f(x) > 3, so the minimum is 3.

Case analysis on a quadratic: when does x² - 5x + 6 > 0?

Answer: x 3

Factor: (x - 2)(x - 3) > 0; both factors positive gives x > 3, both negative gives x < 2.

A function is defined as f(x) = 2 if x is rational, and f(x) = 0 if x is irrational. What is f(π) + f(1/2)?

Answer: 2

π is irrational so f(π) = 0; 1/2 is rational so f(1/2) = 2; sum = 0 + 2 = 2.