Step 1: Determine which piece to use: We need to evaluate f(3) Is 3 < 0? No Is 3 ≥ 0? Yes

Step 2: Use the appropriate formula: For x ≥ 0, use f(x) = 2x - 1

Step 3: Substitute x = 3: f(3) = 2(3) - 1 = 6 - 1 = 5

Answer: f(3) = 5

For $f(-3)$:

Since $-3 < 0$, use the first formula: $f(-3) = 2(-3) = -6$

For $f(2)$:

Since $2 \geq 0$, use the second formula: $f(2) = 2 + 5 = 7$

Answer: $f(-3) = -6$, $f(2) = 7$

Step 1: Evaluate f(-2): Is -2 < 0? Yes Use f(x) = x² f(-2) = (-2)² = 4

Step 2: Evaluate f(0): Is 0 < 0? No Is 0 ≥ 0? Yes Use f(x) = x + 1 f(0) = 0 + 1 = 1

Answer: f(-2) = 4, f(0) = 1

The absolute value changes behavior at the point where the inside equals zero.

$x - 3 = 0$ when $x = 3$

When $x \geq 3$: the inside is positive or zero $|x - 3| = x - 3$

When $x < 3$: the inside is negative $|x - 3| = -(x - 3) = -x + 3$

Answer: $f(x) = \begin{cases} -x + 3 & \text{if } x < 3 \\ x - 3 & \text{if } x \geq 3 \end{cases}$

Step 1: Graph first piece (x < 1): f(x) = -x for x < 1 This is a line with slope -1 At x = 0: f(0) = 0 At x = 1: f(1) = -1 (open circle, not included)

Step 2: Graph second piece (x ≥ 1): f(x) = x - 2 for x ≥ 1 This is a line with slope 1 At x = 1: f(1) = -1 (closed circle, included) At x = 3: f(3) = 1

Step 3: Identify domain: All real numbers (both pieces cover all x values) Domain: (-∞, ∞)

Step 4: Identify range: First piece goes from 1 down (as x approaches 1 from left) Second piece goes from -1 up Range: (-∞, ∞)

Answer: Domain: all real numbers, Range: all real numbers

Step 1: Recall absolute value definition: |a| = { a, if a ≥ 0 {-a, if a < 0

Step 2: Determine when (x - 2) ≥ 0: x - 2 ≥ 0 x ≥ 2

Step 3: Determine when (x - 2) < 0: x - 2 < 0 x < 2

Step 4: Write piecewise function: For x ≥ 2: |x - 2| = x - 2 For x < 2: |x - 2| = -(x - 2) = -x + 2

Step 5: Write in standard form: f(x) = { -x + 2, if x < 2 { x - 2, if x ≥ 2

Answer: f(x) = { -x + 2, if x < 2 { x - 2, if x ≥ 2

Check if the function value and limits match at $x = 1$.

From left: (use $x^2 + 1$) $\lim_{x \to 1^-} f(x) = 1^2 + 1 = 2$

From right: (use $3x - 1$) $\lim_{x \to 1^+} f(x) = 3(1) - 1 = 2$

Function value: (use $x^2 + 1$ since $1 \leq 1$) $f(1) = 1^2 + 1 = 2$

Since all three equal 2, the function is continuous at $x = 1$.

Answer: Yes, continuous at $x = 1$

Step 1: Define the function C(t) for time t hours: For 0 < t ≤ 1: C(t) = 5 For 1 < t ≤ 5: C(t) = 5 + 3(t - 1) = 3t + 2 For t > 5: C(t) = 20

Step 2: Write complete piecewise function: C(t) = { 5, if 0 < t ≤ 1 { 3t + 2, if 1 < t ≤ 5 { 20, if t > 5

Step 3: Calculate cost for 3.5 hours: 3.5 is in range 1 < t ≤ 5 C(3.5) = 3(3.5) + 2 = 10.5 + 2 = 12.5

Step 4: Calculate cost for 6 hours: 6 > 5 C(6) = 20

Answer: C(t) = { 5, if 0 < t ≤ 1 { 3t + 2, if 1 < t ≤ 5 { 20, if t > 5 C(3.5) = $12.50, C(6) =$20