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Chapter 1: Problem 59

Find the only function whose domain is the set of real numbers and that is both even and odd.

Short Answer

Expert verified
The only function with the domain as the set of real numbers and that is both even and odd is \(f(x) = 0\).

Step by step solution

01

Define even function

An even function is defined as f(-x) = f(x) for all x in its domain. This means that the function remains the same when x is replaced with -x.
02

Define odd function

An odd function is defined as f(-x) = -f(x) for all x in its domain. This means that the function changes its sign when x is replaced with -x.
03

Combine the definitions

Since we are looking for a function that is both even and odd, we need to find a function that meets both definitions. This means that f(-x) should be equal to both f(x) and -f(x) simultaneously.
04

Solve for the function

Based on the definitions, we have the two equations: 1. f(-x) = f(x) 2. f(-x) = -f(x) Now, we will equate the right side of the above equations: f(x) = -f(x) To find the function, we have to add f(x) to both sides of the equation: f(x) + f(x) = 0 2f(x) = 0 Now, we need to find f(x), so we divide both sides of the equation by 2: f(x) = 0
05

Describe the solution

The only function that meets both definitions of even and odd functions is f(x) = 0. This means that the function is constant and equal to 0 for all the real numbers in its domain.

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Most popular questions from this chapter

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: The sum of an even function and an odd function (with the same domain) is an odd function. (b) Explain your answer to part (a). This means that if the answer is "true", then explain why the sum of every even function and every odd function (with the same domain) is an odd function; if the answer is "false", then give an example of an even function \(f\) and an odd function \(g\) (with the same domain) such that \(f+g\) is not an odd 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.

Explain why an even function whose domain contains a nonzero number cannot be a one-to- \(-\) one function.

Suppose h is defined by \(h(t)=|t|+1\). What is the range of \(h\) if the domain of \(h\) is the interval [-3,5]\(?\)
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