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Chapter 3: Problem 29

Graph each function. Be sure to label three points on the graph. If \(f(x)=\left\\{\begin{array}{ll}2 x+4 & \text { if }-3 \leq x \leq 1 \\\ x^{3}-1 & \text { if } 1

Short Answer

Expert verified
f(-2)=0, f(0)=4, f(1)=6, f(3)=26.

Step by step solution

01

Understand the Function Piecewise Definition

The function is defined in two pieces. For -3 \leq x \leq 1 \, the function is given by \(f(x)=2x+4\), and for 1<x \leq 5, \, it is given by \(f(x)=x^3-1\).
02

Evaluate f(-2)

Since -3 \leq -2 \leq 1, we use the first piece of the function. Substitute -2 into \(2x+4\): \(f(-2) = 2(-2)+4=-4+4=0\).
03

Evaluate f(0)

Since -3 \leq 0 \leq 1, we use the first piece of the function. Substitute 0 into \(2x+4\): \(f(0)=2(0)+4=4\).
04

Evaluate f(1)

Since -3 \leq 1 \leq 1, we use the first piece of the function. Substitute 1 into \(2x+4\): \(f(1)=2(1)+4=6\).
05

Evaluate f(3)

Since 1 < 3 \leq 5, we use the second piece of the function. Substitute 3 into \(x^3-1\): \(f(3)=3^3-1=27-1=26\).
06

Graph each Piece of the Function

For -3 \leq x \leq 1, graph \(y=2x+4\). For 1<x \leq 5, graph \(y=x^3-1\). Be sure to label the points (-2,0), (0,4), (1,6), and (3,26) on the graph.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Graphing Functions
When graphing piecewise functions, treat each 'piece' of the function as a separate line or curve. In our problem, for the interval \text{-3 \textbackslash leq x \textbackslash leq 1}, the function is a straight line given by the equation \( y = 2x + 4 \). For the interval \( 1 < x \textbackslash leq 5 \), the function is a cubic curve given by \( y = x^3 - 1 \). The graph of each segment should be plotted within its specific interval. To make your graph more accurate, plot points from the solution: (-2,0), (0,4), (1,6), and (3,26). These points help break the abstraction into tangible values on the graph.
Evaluation of Functions
Evaluating piecewise functions means identifying which 'piece' of the function to use for a particular value of \( x \). Depending on where an \( x \) value falls on the domain, you use a specific formula. For example, for \( x = -2 \), since \( -2 \textbackslash, lies in the interval [-3, 1] \), you use the function \( f(x)=2x+4 \) to find \( f(-2) = 0 \). Similarly, if we need \( f(3) \) which lies in the interval \( 1 < x \textbackslash leq 5 \), we use \( f(x)=x^{3}-1 \) to find \( f(3)=26 \).
Domain and Range
The domain of a piecewise function is the combined interval over which each individual piece is defined. In this problem, the domain is \( -3 \leq x \textbackslash leq 5 \). Each 'piece' of the function also has its own range. For \( f(x) = 2x + 4 \) from \( -3 \leq x \textbackslash leq 1 \), you calculate the values at the endpoints of this interval to get a range of \( -2 \leq f(x)\textbackslash leq 6\). For \( f(x) = x^3 - 1 \) from \( 1 < x \textbackslash leq 5 \), its range is calculated by plugging in the endpoints to find \( -0 \textbackslash leq f(x) \leq 124\). Together, they give a total range that includes all values these two ranges cover.

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Most popular questions from this chapter

Use a graphing utility to graph each function over the indicated interval and approximate any local maximum values and local minimum values. Determine where the function is increasing and where it is decreasing. Round answers to two decimal places. \(f(x)=0.25 x^{4}+0.3 x^{3}-0.9 x^{2}+3 \quad[-3,2]\)

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