1-3 Additional Practice Piecewise Defined Functions Answer Key

Mastering Piecewise Defined Functions: 3 Additional Practice Problems with Detailed Solutions

Piecewise defined functions, those intriguing mathematical creatures with multiple expressions defined across different intervals, often present a challenge for students. Understanding how to evaluate, graph, and manipulate these functions is crucial for success in calculus and beyond. This article provides three additional practice problems, each with a detailed step-by-step solution, to solidify your understanding of piecewise functions. We'll cover everything from evaluating the function at specific points to graphing the function and analyzing its properties.

Problem 1: A Multi-Part Function with Absolute Value

Let's consider the following piecewise function:

f(x) =  { x² + 1,  if x < -1
         { |x|,      if -1 ≤ x ≤ 2
         { 3x - 2,  if x > 2

A. Evaluate the function at the following points:

1. f(-2)
2. f(-1)
3. f(0)
4. f(2)
5. f(3)

B. Sketch the graph of f(x).

C. Identify the domain and range of f(x).

Solution:

A. Evaluating the function:

1. f(-2): Since -2 < -1, we use the first expression: f(-2) = (-2)² + 1 = 4 + 1 = 5

2. f(-1): Since -1 ≤ -1 ≤ 2, we use the second expression: f(-1) = |-1| = 1

3. f(0): Since -1 ≤ 0 ≤ 2, we use the second expression: f(0) = |0| = 0

4. f(2): Since -1 ≤ 2 ≤ 2, we use the second expression: f(2) = |2| = 2

5. f(3): Since 3 > 2, we use the third expression: f(3) = 3(3) - 2 = 9 - 2 = 7

B. Sketching the graph:

To sketch the graph, we consider each part separately.

• x < -1: This is a parabola, y = x² + 1, but only for the portion where x is less than -1. The point (-1, 2) is an open circle because x = -1 is not included in this interval.

• -1 ≤ x ≤ 2: This is the absolute value function, y = |x|, but only from x = -1 to x = 2. The points (-1, 1) and (2, 2) are closed circles because these x-values are included.

• x > 2: This is a straight line, y = 3x - 2, starting from x = 2. The point (2, 4) is an open circle because x =2 is not included in this interval. However, since f(2) is defined as 2 from the previous part, we have a closed circle at (2,2).

By plotting these points and connecting them according to the defined intervals, you obtain the graph of the piecewise function.

C. Domain and Range:

• Domain: The domain is all real numbers, (-∞, ∞), since the function is defined for all x-values.

• Range: Observing the graph, the range is [0, ∞). The function values never go below 0.

Problem 2: A Piecewise Function with a Step Function Component

Consider the function:

g(x) = { ⌊x⌋,       if x < 0
        { x² - 1,   if 0 ≤ x ≤ 3
        { 8,         if x > 3

where ⌊x⌋ represents the greatest integer less than or equal to x (the floor function).

A. Evaluate:

1. g(-2.5)
2. g(0)
3. g(1.8)
4. g(3)
5. g(4)

B. Graph g(x).

C. State the domain and range of g(x).

Solution:

A. Evaluating the function:

1. g(-2.5): Since -2.5 < 0, we use the floor function: g(-2.5) = ⌊-2.5⌋ = -3

2. g(0): Since 0 ≤ 0 ≤ 3, we use the second expression: g(0) = 0² - 1 = -1

3. g(1.8): Since 0 ≤ 1.8 ≤ 3, we use the second expression: g(1.8) = (1.8)² - 1 = 3.24 - 1 = 2.24

4. g(3): Since 0 ≤ 3 ≤ 3, we use the second expression: g(3) = 3² - 1 = 8

5. g(4): Since 4 > 3, we use the third expression: g(4) = 8

B. Sketching the graph:

• x < 0: This is a step function. For each interval [n, n+1) where n is an integer, the function value is n.

• 0 ≤ x ≤ 3: This is a parabola, y = x² - 1, defined on the interval [0, 3].

• x > 3: This is a horizontal line at y = 8.

Combining these parts, you'll notice a discontinuity at x = 0 and a continuous connection at x = 3.

C. Domain and Range:

• Domain: The domain is all real numbers, (-∞, ∞).

• Range: The range is a bit more complex due to the step function and parabola. The range encompasses all integers less than 0, combined with the values of the parabola from -1 to 8, and finally the constant value 8. Therefore the range is (-∞, 0) U [-1, 8]

Problem 3: A Piecewise Function Involving Rational Expressions

Let's analyze the function:

h(x) = { (x+2)/(x-1),  if x < -2
        { 1,            if -2 ≤ x ≤ 1
        { x²/(x-2),     if x > 1 and x≠2

A. Find h(-3), h(-2), h(0), h(1), h(3), and h(2).

B. Describe any discontinuities in h(x).

C. Determine the domain and range of h(x).

Solution:

A. Evaluating the function:

1. h(-3): Since -3 < -2, we use the first expression: h(-3) = (-3 + 2)/(-3 - 1) = -1/-4 = 1/4

2. h(-2): Since -2 ≤ -2 ≤ 1, we use the second expression: h(-2) = 1

3. h(0): Since -2 ≤ 0 ≤ 1, we use the second expression: h(0) = 1

4. h(1): Since -2 ≤ 1 ≤ 1, we use the second expression: h(1) = 1

5. h(3): Since 3 > 1 and 3 ≠ 2, we use the third expression: h(3) = 3²/(3 - 2) = 9/1 = 9

6. h(2): The third expression is undefined at x = 2 because the denominator is zero. Therefore, h(2) is undefined.

B. Discontinuities:

There are several points of discontinuity.

• x = 1: There's a jump discontinuity at x = 1 because the limit as x approaches 1 from the left is 1, but the limit as x approaches 1 from the right is 1 as well, but because of the value at x=1, this forms a jump discontinuity.

• x = 2: There's an infinite discontinuity (a vertical asymptote) at x = 2 because the third expression is undefined at this point.

C. Domain and Range:

• Domain: The domain is all real numbers except x = 2: (-∞, 2) U (2, ∞).

• Range: Determining the precise range requires a more detailed analysis. The range includes the value 1 and extends to positive and negative infinity due to the rational expressions. Further analysis of the rational parts is needed for a more precise range.

These three problems illustrate the diverse nature of piecewise functions and the techniques needed to handle them. Remember to always carefully consider the intervals and corresponding expressions when evaluating, graphing, and analyzing piecewise defined functions. Practice is key to mastering these functions, so try creating your own piecewise functions and testing your understanding!