MPT Case Analysis 3

Question 1

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

Accepted Answer

x = 1 only

Answer

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) ✓.

Question 2

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

Accepted Answer

The absolute value function |x|

Answer

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

Question 3

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

Accepted Answer

x < -3 or x > 3

Answer

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

Question 4

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?

Accepted Answer

$50

Answer

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

Question 5

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

Accepted Answer

3

Answer

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

(MPT) UCF Math Placement Test Practice Test