Case Analysis Flashcards

A function f is defined as f(x) = 1/(x - 2) for x > 2, and f(x) = x - 5 for x ≤ 2. What is f(2) + f(4)?

Answer: -2.5

f(2) = 2 - 5 = -3 (x ≤ 2 piece); f(4) = 1/(4-2) = 0.5 (x > 2 piece); sum = -3 + 0.5 = -2.5.

Case analysis on absolute value: solve |5 - 2x| ≤ 3.

Answer: 1 ≤ x ≤ 4

-3 ≤ 5 - 2x ≤ 3 → -8 ≤ -2x ≤ -2 → dividing by -2 and flipping: 1 ≤ x ≤ 4.

A piecewise function has f(x) = 4 for all x. Which statement correctly describes its case structure?

Answer: It can be written as piecewise with identical pieces on each region

Any function can be written piecewise; a constant function simply uses the same value in every case, though this adds no information.

A savings plan adds $100/month for balances below $1,000 and $150/month for balances ≥ $1,000. Starting at $950, what is the balance after 2 months?

Answer: $1,200

Month 1: $950 < $1,000 → add $100 → $1,050; Month 2: $1,050 ≥ $1,000 → add $150 → $1,200.

For f(x) = x + 1 if x < 1 and f(x) = x² if x ≥ 1, is f continuous at x = 1?

Answer: No, because the left limit is 2 but f(1) = 1

Left limit: lim(x→1⁻) x+1 = 2; actual value: f(1) = 1² = 1; since 2 ≠ 1, f is not continuous at x = 1.

Using case analysis, simplify √(x²) for all real x.

Answer: |x|

√(x²) = |x| because for x ≥ 0 it equals x, and for x < 0 it equals -x, which together define the absolute value.

A piecewise function f(x) = ax + 1 for x < 2 and f(x) = x² - 1 for x ≥ 2. For what value of a is f continuous at x = 2?

Answer: a = 1

Right-side value: f(2) = 2² - 1 = 3; left-side limit: 2a + 1 = 3 → 2a = 2 → a = 1.