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Ch. 1 - Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 1 - FunctionsProblem 31
Chapter 1, Problem 31

Piecewise linear functions Graph the following functions.
f(x)={3x−1, if x≤0−2x−1, if x>0f\(\left\)(x\(\right\))=\(\begin{cases}\)3x-1\(\frac{}{}\),\(\text{ if }\)x\(\le\)0\\ -2x-1,\(\text{ if }\)x>0\(\end{cases}\)

Verified step by step guidance
1
Step 1: Understand the piecewise function definition. The function \( f(x) \) is defined as \( 3x - 1 \) for \( x \leq 0 \) and \( -2x - 1 \) for \( x > 0 \).
Step 2: Graph the first piece of the function, \( 3x - 1 \), for \( x \leq 0 \). This is a linear function with a slope of 3 and a y-intercept of -1. Plot the line starting from the y-intercept at (0, -1) and extend it to the left.
Step 3: Graph the second piece of the function, \( -2x - 1 \), for \( x > 0 \). This is a linear function with a slope of -2 and a y-intercept of -1. Plot the line starting from the y-intercept at (0, -1) and extend it to the right.
Step 4: Consider the point where the two pieces meet at \( x = 0 \). For \( x = 0 \), the value from the first piece is \( 3(0) - 1 = -1 \). Since the condition is \( x \leq 0 \), the point (0, -1) is included in the graph.
Step 5: Combine the two pieces on the graph. The graph will have a solid point at (0, -1) for the first piece and an open point at (0, -1) for the second piece, indicating that the second piece does not include \( x = 0 \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Piecewise Functions

Piecewise functions are defined by different expressions based on the input value. In this case, the function f(x) has two distinct linear equations: one for x less than or equal to 0, and another for x greater than 0. Understanding how to interpret and graph these functions requires recognizing the conditions under which each piece applies.
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Piecewise Functions

Graphing Linear Functions

Graphing linear functions involves plotting points that satisfy the linear equations and connecting them to form straight lines. Each piece of the piecewise function can be graphed separately, taking care to indicate where each piece is valid. The slopes and intercepts of the lines are crucial for accurately representing the function on a coordinate plane.
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Linearization

Continuity and Discontinuity

Continuity refers to whether a function is unbroken at a point, while discontinuity indicates a 'jump' or break in the graph. For piecewise functions, it is essential to check the endpoints where the pieces meet, as this can affect the overall behavior of the function. In this case, examining the function at x = 0 will reveal if it is continuous or if there is a jump between the two linear pieces.
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Intro to Continuity
Related Practice
Textbook Question

Taxicab fees A taxicab ride costs \$3.50 plus \$2.50 per mile for the first 5 miles, with the rate dropping to \$1.50 per mile after the fifth mile. Let m be the distance (in miles) from the airport to a hotel. Find and graph the piecewise linear function c(m) that represents the cost of taking a taxi from the airport to a hotel m miles away.

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Textbook Question

Find the inverse function (on the given interval, if specified) and graph both ff and f−1f^{-1} on the same set of axes. Check your work by looking for the required symmetry in the graphs.

f(x)=x2+4f\(\left\)(x\(\right\))=x^2+4, for x≥0x\(\geq{0}\)

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Textbook Question

Graphing functions Sketch a graph of each function.


ƒ(x) = { 2x if x ≤ 1 , 3-x if x > 1

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Textbook Question

{Use of Tech} Launching a rocket A small rocket is launched vertically upward from the edge of a cliff 8080 ft above the ground at a speed of 9696 ft/s. Its height (in feet) above the ground is given by h(t)=−16t2+96t+80h\(\left\)(t\(\right\))=-16t^2+96t+80, where tt represents time measured in seconds.

a. Assuming the rocket is launched at t=0t=0, what is an appropriate domain for hh?

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Textbook Question

Draining a tank (Torricelli’s law) A cylindrical tank with a cross-sectional area of 1010 m2 is filled to a depth of 2525 m with water. At t=0t=0 s, a drain in the bottom of the tank with an area of 11 m²  is opened, allowing water to flow out of the tank. The depth of water in the tank (in meters) at time t≥0t\(\geq{0}\) is d(t)=(5−0.22t)2d\(\left\)(t\(\right\))=\(\left\)(5-0.22t\(\right\))^2.

a. Check that d(0)=25d\(\left\)(0\(\right\))=25, as specified.

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Textbook Question

Graphing equations Graph the following equations. 


c. x² + 2x + y² + 4y + 1 = 0

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