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Chapter 1: Problem 57

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

Short Answer

Expert verified
To prove that the product of two even functions is also an even function, let's denote the given functions as \(f(x)\) and \(g(x)\), both satisfying the condition \(f(x) = f(-x)\) and \(g(x) = g(-x)\) for all x in their domain. Let h(x) be the product of these functions, so \(h(x) = f(x) * g(x)\). We now substitute -x for x and get \(h(-x) = f(-x) * g(-x)\). Using the properties of even functions, we have \(h(-x) = f(x) * g(x)\), which means \(h(-x) = h(x)\), proving that the product function h(x) is also even.

Step by step solution

01

Define the given functions

Let's denote the two given even functions as f(x) and g(x), where f(x) = f(-x) and g(x) = g(-x) for all x in their domain.
02

Find the product of the functions

To reveal that the product of the functions is even as well, we first need to express their product. Let h(x) be their product, so h(x) = f(x) * g(x) for all x in their domain.
03

Substitute with the negative argument

Now, we will substitute x with -x in the expression for h(x), and show that h(-x) = h(x) for the product of f(x) and g(x). This will prove that theproduct is also an even function. h(-x) = f(-x) * g(-x)
04

Use the properties of even functions

Given that f(x) and g(x) are even functions, we can use their respective properties of evenness here: h(-x) = f(x) * g(x) This means that h(-x) = h(x) which implies that the product of the two even functions, h(x), is an even function as well.
05

Conclusion

We have demonstrated that the product of two even functions, f(x) and g(x), is also an even function, since h(-x) = h(x) for h(x) = f(x) * g(x). Therefore, the product of two even functions with the same domain is an even function.

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Most popular questions from this chapter

For Exercises \(47-50,\) suppose \(f\) is a function whose domain is the interval [-5,5] and that $$ f(x)=\frac{x}{x+3} $$ for every \(x\) in the interval [0,5] . Suppose \(f\) is an even function. Evaluate \(f(-2)\).

Give an example of a function whose domain is the set of positive integers and whose range is the set of integers.

Show that the composition of two one-to-one functions is a one-to-one function. [Here you need to assume that the two functions have range and domain such that their composition makes sense.]

(a) True or false: Just as every integer is either even or odd, every function whose domain is the set of integers is either an even function or an odd function. (b) Explain your answer to part (a). This means that if the answer is "true", then you should explain why every function whose domain is the set of integers is either an even function or an odd function; if the answer is "false", then you should give an example of a function whose domain is the set of integers but that is neither even nor odd.

Give an example of a function whose domain is the set of positive even integers and whose range is the set of positive odd integers.
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