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Chapter 1: Problem 37

Suppose \(f\) is the function whose domain is the interval \([-2,2],\) with \(f\) defined by the following formula: $$f(x)=\left\\{\begin{array}{ll}-\frac{x}{3} & \text { if }-2 \leq x<0 \\\2 x & \text { if } 0 \leq x \leq 2\end{array}\right.$$ (a) Sketch the graph of \(f\). (b) Explain why the graph of \(f\) shows that \(f\) is not a one-to-one function. (c) Give an explicit example of two distinct numbers \(a\) and \(b\) such that \(f(a)=f(b) .\)

Short Answer

Expert verified
(a) The graph of \(f(x)\) is composed of two straight lines. For the interval \([-2, 0)\), the function is given by \(f(x) = -\frac{x}{3}\), and for the interval \([0, 2]\), the function is given by \(f(x) = 2x\). (b) The graph of \(f(x)\) shows that there are horizontal lines that intersect the graph at more than one point, therefore it is not a one-to-one function. (c) An explicit example of two distinct numbers a and b with the same function value is: \(a = -1\) and \(b = \frac{1}{6}\) such that \(f(a) = f(b) = \frac{1}{3}\).

Step by step solution

01

Part (a): Sketching the graph of \(f(x)\)

To sketch the graph, we will use the two given formulas for different intervals of the domain: 1. For the interval \([-2, 0)\), the function is given by \(f(x) = -\frac{x}{3}\). 2. For the interval \([0, 2]\), the function is given by \(f(x) = 2x\). First, note that the function is continuous. To sketch the graph, we will plot points and join them with straight lines. For the interval \([-2, 0)\): - At \(x=-2\), we have \(f(-2) = -\frac{-2}{3} = \frac{2}{3}\). - At \(x=-1\), we have \(f(-1)=-\frac{-1}{3}=\frac{1}{3}\). - At \(x=-0.5\), we have \(f(-0.5)=-\frac{-0.5}{3} = \frac{1}{6}\). For the interval \([0, 2]\): - At \(x=0\), we have \(f(0) = 2\cdot 0 = 0\). - At \(x=1\), we have \(f(1) = 2\cdot 1 = 2\). - At \(x=2\), we have \(f(2)= 2\cdot 2 = 4\). Now, plot these points on a graph and join them with straight lines because these are linear equations for both intervals. This will give the graph of \(f(x)\) in the domain\([-2, 2]\).
02

Part (b): Determining one-to-one property

A function is one-to-one if, for any distinct pair of inputs (\(x_1\) and \(x_2)\), their corresponding outputs are also distinct (\(f(x_1) \neq f(x_2)\)). In other words, a function is one-to-one if it never takes the same value twice. To determine if the function is one-to-one, we can look at its graph. Take any horizontal line, and if it intersects the graph of the function more than once, the function is not one-to-one. In our graph, the function is not one-to-one because there are horizontal lines that intersect the graph at more than one point. For example, take the horizontal line \(y=\frac{1}{3}\). This line intersects the graph at two points.
03

Part (c): Providing an explicit example

We need to find two distinct numbers a and b in the domain [-2, 2] that have the same function value. From the graph, we can observe that the point \((x, f(x)) = (-1, \frac{1}{3})\) on the first part of the function satisfies this condition. To find the corresponding point or value on the second part of the function, we equate \(f(x) = \frac{1}{3}\) in its definition for the \(0 \leq x \leq 2\) interval. That is, \(2x = \frac{1}{3}\). Solving for x, we get \(x = \frac{1}{6}\). So, our point is \((x, f(x)) = \left(\frac{1}{6}, \frac{1}{3}\right)\). Thus, an explicit example of two distinct numbers a and b with the same function value is: \(a = -1\) and \(b = \frac{1}{6}\) such that \(f(a) = f(b) = \frac{1}{3}\).

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Key Concepts

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

Domain and Range
A function's domain refers to the set of all possible input values (usually represented as "x") for which the function is defined. In the given problem, the domain is the interval \([-2, 2]\), which means that the function can only accept values of \(x\) ranging from \(-2\) to \(2\), inclusive. Each piece of the piecewise function uses different rules, depending on the interval within this domain.

The range of a function consists of all possible output values (or "y" values) that result from using the domain. To determine the range, we look at each piece of the piecewise function:

  • For \(-2 \leq x < 0\), the output is given by the equation \(f(x) = -\frac{x}{3}\). Substituting \(-2\) gives \(\frac{2}{3}\) and nearby to \(0\) yields \(\approx 0\). Hence, the range for this piece is \(\left[0, \frac{2}{3}\right)\).
  • For \(0 \leq x \leq 2\), the output follows \(f(x) = 2x\), starting at \(0\) and ending at \(4\), giving a range of \([0, 4]\).
When combining these two intervals, we have the entire range for the function \([0, 4]\) as the function outputs all values from \(0\) to \(4\) across both segments.
One-to-One Functions
A one-to-one function is a type of function where each output is mapped to one unique input, which means no two different inputs have the same output. This is sometimes referred to as an "injective" function. Determining whether a function is one-to-one involves checking that no horizontal line intersects the graph of the function more than once.

For the function \(f\) in our exercise, it is not one-to-one because it鈥檚 possible to find different \(x\) values that produce the same \(f(x)\) value. For instance, with \(f(x) = \frac{1}{3}\), this value is achieved at both \(x = -1\) on one interval and \(x = \frac{1}{6}\) on another interval, which is shown in the graph of \(f\).

The intersection of a horizontal line like \(y = \frac{1}{3}\) with the graph at two points (\(-1\) and \(\frac{1}{6}\)) confirms that multiple inputs map to the same output, demonstrating \(f\) is not one-to-one.
Graphing Functions
Graphing a piecewise function involves plotting each piece according to its domain and connecting these points smoothly, given that the function may or may not be continuous. In this case, it's crucial to understand how to precisely sketch each segment outlined by their specific algebraic expressions.

For the given function \(f\), it is constructed of two linear equations. The graph is drawn in the following steps:

  • For the segment defined by \(-2 \leq x < 0\), plot the points resulting from \(f(x) = -\frac{x}{3}\). This gives a set of points like \((-2, \frac{2}{3})\), \((-1, \frac{1}{3})\), and approximations near \(0\).
  • For \(0 \leq x \leq 2\), plot using \(f(x) = 2x\). Key points include \((0, 0)\), \((1, 2)\), and \((2, 4)\).

In each case, you connect these plotted points with straight lines because both expressions describe linear dependencies. Remember, a piecewise function might not be differentiable or continuous everywhere, so it's necessary to check the behavior at boundary points, which in this function happen to align precisely at \(x = 0\), where both lines would meet on the graph if extended.

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Most popular questions from this chapter

Find functions \(f\) and \(g\), each simpler than the given function \(h\), such that \(h=f \circ g\). \(h(x)=\left(x^{2}-1\right)^{2}\)

Suppose \(f\) and \(g\) are functions, each with domain of four numbers, with \(f\) and \(g\) defined by the tables below: $$\begin{array}{c|c}x & f(x) \\\\\hline 1 & 4 \\\2 & 5 \\\3 & 2 \\\4 & 3\end{array}$$ $$\begin{array}{c|c}x & g(x) \\\\\hline 2 & 3 \\\3 & 2 \\\4 & 4 \\\5 & 1\end{array}$$ Give the table of values for \(f \circ f^{-1}\).

Check your answer by evaluating the appropriate function at your answer. Suppose \(f(x)=7 x-5\). Evaluate \(f^{-1}(-3)\).

Suppose you are exchanging cur. rency in the London airport. The currency exchange service there only makes transactions in which one of the two currencies is British pounds, but you want to exchange dollars for Euros. Thus you first need to exchange dollars for British pounds, then exchange British pounds for Euros. At the time you want to make the exchange, the function \(f\) for exchanging dollars for British pounds is given by the formula \(f(d)=0.66 d-1\) and the function \(g\) for exchanging British pounds for Euros is given by the formula \(g(p)=1.23 p-2\) The subtraction of 1 or 2 in the number of British pounds or Euros that you receive is the fee charged by the currency exchange service for each transaction. How many Euros would you receive for exchanging 200 after going through this two-step exchange process?

Find functions \(f\) and \(g\), each simpler than the given function \(h\), such that \(h=f \circ g\). \( h(x)=\frac{2}{3+\sqrt{1+x}}\)
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