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Chapter 1: Problem 64

True or false: The product of an even function and an odd function (with the same domain) is an odd function. Explain your answer.

Short Answer

Expert verified
True, the product of an even function and an odd function (with the same domain) is an odd function. This is because when we calculate the product function \(h(x) = f(x)g(x)\), and evaluate it at \(-x\), we get \(h(-x) = f(-x)g(-x) = f(x)(-g(x)) = -f(x)g(x) = -h(x)\), demonstrating that the product function has the property of an odd function.

Step by step solution

01

Review the properties of even and odd functions

An even function, \(f(x)\), has the following property: \(f(-x) = f(x)\) for all values of \(x\) in its domain. An odd function, \(g(x)\), has the following property: \(g(-x) = -g(x)\) for all values of \(x\) in its domain.
02

Define the product function

Let's define the product function, \(h(x)\), as the product of an even function, \(f(x)\), and an odd function, \(g(x)\): \(h(x) = f(x)g(x)\).
03

Test the product function for even or odd properties

Now, we will check if the product function, \(h(x)\), is an odd function by evaluating the expression for \(h(-x)\), and compare it with the expression for \(h(x)\). \(h(-x) = f(-x)g(-x)\) Since \(f(x)\) is even, we can replace \(f(-x)\) with \(f(x)\); and since \(g(x)\) is odd, we can replace \(g(-x)\) with \(-g(x)\). \(h(-x) = f(x) \cdot (-g(x)) = -f(x)g(x)\) Notice that the expression -\(f(x) \cdot g(x)\) is equivalent to \(-h(x)\). Therefore, the product function has the property \(h(-x) = -h(x)\), which characterizes odd functions.
04

Conclusion

True, the product of an even function and an odd function (with the same domain) is an odd function.

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Most popular questions from this chapter

Suppose \(f\) is a function and a function \(g\) is defined by the given expression. (a) Write \(g\) as the composition of \(f\) and one or two linear functions. (b) Describe how the graph of \(g\) is obtained from the graph of \(f\). \( g(x)=f\left(-\frac{2}{3} x\right)\)

Show that the product of two even functions (with the same domain) is an even function.

Give an example to show that the sum of two one-to-one functions is not necessarily a one-to-one function.

Suppose \(f\) is a function whose domain equals \\{2,4,7,8,9\\} and whose range equals \(\\{-3,0,2,6,7\\} .\) Explain why \(f\) is a one-to-one function.

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. Is the function describing the exchange of dollars for Euros \(f \circ g\) or \(g \circ f ?\) Explain your answer in terms of which function is evaluated first when computing a value for a composition (the function on the left or the function on the right?).
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