Math 2 Piecewise Functions Worksheet 2 Answers

Math 2 Piecewise Functions Worksheet 2 Answers
Mastering piecewise functions is a crucial step in Algebra 2 and prepares students for calculus, modeling real‑world situations, and standardized tests. This article walks through the concepts behind Math 2 Piecewise Functions Worksheet 2, provides a clear strategy for tackling each problem, and offers a detailed answer key with explanations so you can check your work, understand where you went wrong, and build confidence for future assignments.

Introduction

Piecewise functions are functions defined by multiple sub‑functions, each applying to a specific interval of the domain. In Algebra 2, students learn to graph these functions, evaluate them at given x‑values, write equations from graphs, and solve real‑world problems that change behavior at certain thresholds (e.g., tax brackets, shipping rates, or cell‑phone plans). Worksheet 2 typically focuses on:

• Identifying the correct sub‑function for a given x.
• Graphing each piece accurately, including open and closed endpoints. * Determining domain and range from a piecewise definition.
• Writing a piecewise rule from a verbal description or a graph.

The following sections break down the worksheet into manageable steps, highlight common pitfalls, and provide a full set of solutions with reasoning.

Understanding Piecewise Functions

Before diving into the worksheet, refresh the core definition:

A piecewise function ( f(x) ) is expressed as > [ f(x)=\begin{cases} f_1(x), & \text{if } x\in I_1\[4pt] f_2(x), & \text{if } x\in I_2\[4pt] \vdots & \ f_n(x), & \text{if } x\in I_n> \end{cases} ] where each ( I_k ) is an interval (often expressed with inequalities) and the intervals partition the domain without overlap (except possibly at boundary points).

Key points to remember:

• Endpoint notation matters: a closed circle (•) means the point is included (≤ or ≥); an open circle (∘) means it is excluded (< or >).
• When graphing, treat each piece independently, then combine them, respecting the endpoint rules.
• The domain is the union of all intervals; the range is found by evaluating the output values over those intervals.

How to Use the Worksheet

Worksheet 2 is designed to reinforce the above ideas through a mix of short‑answer, graphing, and word‑problem items. Follow this workflow for each question:

1. Read the prompt carefully – identify whether you are given a rule, a graph, or a description.
2. Highlight the intervals – write down the inequality that defines each piece.
3. Choose the appropriate sub‑function – plug the given x into the correct formula, or sketch the corresponding graph segment.
4. Check endpoints – decide if the point should be plotted as a closed or open circle.
5. Verify your answer – for graphing questions, use a test point inside each interval to confirm the shape; for algebraic questions, substitute back into the original piecewise definition.

Step‑by‑Step Guide to Solving Problems

Below is a generic step‑by‑step method that applies to most items on Worksheet 2. Adjust the details according to the specific problem type.

1. Identify the Type of Question

2. Extract the Intervals

• Write each interval as an inequality (e.g., ( -3 \le x < 2 )).
• If the worksheet gives a graph, read the x‑coordinates where the graph changes shape or where a hole/jump appears.

3. Match the Sub‑Function

• For evaluation: locate which inequality contains the input value, then substitute into the corresponding expression.
• For graphing: sketch each sub‑function on its interval, using a light pencil line first, then darken the portion that belongs to the piece.
• For rule writing: observe the behavior on each interval, determine the underlying function (linear, quadratic, constant, etc.), and note whether endpoints are included.

4. Apply Endpoint Rules

• Closed circle → include the point (use ≤ or ≥).
• Open circle → exclude the point (use < or >). * When writing the rule, reflect this by using the appropriate inequality symbol.

5. State Domain and Range (if required) * Domain = union of all intervals (often written as ([a,b]\cup[c,d]) etc.).

• Range = evaluate the minimum and maximum y‑values each piece can attain on its interval, then combine.

6. Verify

• Plug a few test x‑values from each interval into your final rule and compare with the original graph or description.
• For graphing, ensure that the picture matches the described behavior at every boundary. ---

Common Types of Problems on Worksheet 2

A. Evaluation Problems Example: Given

[ f(x)=\begin{cases} 2x+1, & x<0\ x^2-4, & x\ge 0 \end{cases} ]
Find ( f(-3) ) and ( f(2) ).

Solution:

• For ( x=-3 ), the condition ( x<0 ) applies → ( f(-3)=2(-3)+1=-5 ).
• For ( x=2 ), the condition ( x\ge0 ) applies → ( f(2)=2^2-4=0 ).

B. Graphing Problems

Example: Graph
[ g(x)=\begin{cases} -x+2, & -2\le x<1\ 3, & x=1\ 0.5x-1, & x>1\end{cases} ]

Solution Sketch:

1. Plot the line ( -x+2 ) from ( x=-2 ) (

C. Rule‑Writing Problems

Example: The graph below consists of three distinct pieces: * a line passing through (‑4, 0) and (0, 4) for (x\le 0);

• a horizontal segment at (y=‑2) from (x=0) to (x=3) (including the left endpoint but not the right);
• a parabola opening upward with vertex at (3, 1) for (x\ge 3).

Task: Write a piecewise definition for the function (h(x)).

Solution:

1. Identify intervals – from the description we have

   • (x\le 0)
   • (0<x<3) * (x\ge 3).

2. Find the sub‑function on each interval

   • For (x\le0): slope (=\frac{4-0}{0-(-4)}=1) and y‑intercept = 4 → (h(x)=x+4).
   • For (0<x<3): constant (y=-2) → (h(x)=-2).
   • For (x\ge3): vertex form of a parabola (y=a(x-3)^2+1). Using the point (4, 2) (read from the graph) gives (2=a(1)^2+1\Rightarrow a=1). Hence (h(x)=(x-3)^2+1).

3. Apply endpoint symbols – the graph shows a closed circle at (‑4, 0) and at (0, 4) (so include (x=0) in the first piece), an open circle at (3, ‑2) (exclude (x=3) from the second piece), and a closed circle at (3, 1) (include (x=3) in the third piece).

Putting it together:

[ h(x)=\begin{cases} x+4, & x\le 0\[4pt] -2, & 0<x<3\[4pt] (x-3)^2+1, & x\ge 3 \end{cases} ]

D. Domain and Range Problems

Example:
Given

[ p(x)=\begin{cases} \displaystyle \frac{1}{x+2}, & x<-2\[6pt] \sqrt{x}, & 0\le x\le 9\[6pt] 5, & x>9 \end{cases} ]

find the domain and range of (p).

Solution:

1. Domain – examine each piece for restrictions.

   • (\frac{1}{x+2}) is undefined when (x+2=0\Rightarrow x=-2); but this piece only applies for (x<-2), so (-2) is already excluded.
   • (\sqrt{x}) requires (x\ge0); the interval given is (0\le x\le9), which satisfies the condition.
   • The constant piece imposes no further restriction. Hence the domain is the union of the intervals:
     [ (-\infty,-2);\cup;[0,9];\cup;(9,\infty)=(-\infty,-2)\cup[0,\infty). ]

2. Range – evaluate the possible y‑values on each interval.

   • For (x<-2), let (u=x+2<0); then (\frac{1}{u}) takes all negative values (approaches (0^{-}) as (x\to-\infty) and (-\infty) as (x\to-2^{-})). So this piece contributes ((-\infty,0)).
   • For (0\le x\le9), (\sqrt{x}) runs from (\sqrt{0}=0) to (\sqrt{9}=3). Including the endpoint gives ([0,3]).
   • For (x>9), the constant value is (5).

   Combining: ((-\infty,0)\cup[0,3]\cup{5}=(-\infty,3]\cup{5}).

Thus

[ \text{Domain}=(-\infty,-2)\cup[0,\infty),\qquad \text{Range}=(-\infty,3]\cup{5}. ]

E. Application Problems

Example:
A taxi company charges $3.00 for the first mile or fraction thereof, then $2.00 for each additional mile or fraction thereof, with a maximum fare of $15.00 for any trip. Write a

...function to model the cost of a taxi ride, (C(x)), where (x) is the number of miles traveled. Determine the domain and range of (C(x)).

Solution:

1. Define the function – The problem describes a piecewise function. We need to break down the cost into segments based on the number of miles traveled. Let's define the cost for each mile traveled.

   • Cost for the first mile or fraction thereof: $3.00
   • Cost for each additional mile or fraction thereof: $2.00

2. Determine the function pieces – We can express the cost as follows:

   • If (0<x\le 1), then (C(x)=3).
   • If (1<x\le 2), then (C(x)=3 + 2(x-1)).
   • If (2<x\le 3), then (C(x)=3 + 2(x-1) + 2(x-2)).
   • And so on...

3. Generalize the function – The general formula for the cost (C(x)) can be written as:

   [ C(x)=\begin{cases} 3, & 0<x\le 1\[4pt] 3+2(x-1), & 1<x\le 2\[4pt] 3+2(x-1)+2(x-2), & 2<x\le 3\[4pt] 3+2(x-1)+2(x-2)+2(x-3), & 3<x\le 4\[4pt] \vdots & \vdots
   \end{cases} ]

   This pattern continues. For each additional mile, we add 2 times the number of miles traveled since the last mile. The cost is effectively the initial fare plus a linear increase in cost for each subsequent mile. We can express this more compactly as:

   [ C(x) = 3 + 2\sum_{i=0}^{x-1} i = 3 + 2 \cdot \frac{(x-1)(x)}{2} = 3 + x(x-1) = x^2 - x + 3 ]

   This formula holds for (x) such that (x^2 - x + 3 \le 15), which is (x^2 - x - 12 \le 0). Factoring gives ((x-4)(x+3) \le 0), so (-3 \le x \le 4). Since (x) represents the number of miles traveled, it must be non-negative, so (0 \le x \le 4). The function (C(x) = x^2 - x + 3) is a parabola opening upwards, with a minimum at (x=1/2). The maximum value occurs at (x=4), which is (C(4) = 4^2 - 4 + 3 = 16 - 4 + 3 = 15).

4. Domain – The domain is the set of all possible values of (x). Since the taxi company has a maximum fare of $15.00, we have (C(x) \le 15). Also, (x) must be non-negative, i.e., (x \ge 0). Therefore, the domain is (0 \le x \le 4).

5. Range – The range is the set of all possible values of (C(x)). Since (C(x) = x^2 - x + 3), we know that the minimum value of (C(x)) occurs at (x=1/2), where (C(1/2) = (1/2)^2 - (1/2) + 3 = 1/4 - 1/2 + 3 = 1/4 - 2/4 + 12/4 = 11/4 = 2.75). The maximum value occurs at (x=4), where (C(4) = 15). Therefore, the range is ([2.75, 15]).

Conclusion:

The function (C(x) = x^2 - x + 3) models the cost of a taxi ride. The domain of (C(x)) is ([0, 4]), and the range is ([2.75, 15]). This indicates that the cost of a taxi ride is between $2.75 and $15.00, and the cost is always increasing as the number of miles traveled increases. The function's behavior reflects the tiered pricing structure of the taxi company, where the initial fare is fixed, and subsequent miles are charged at a decreasing rate.