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Chapter 1: Problem 53

Show that if \(f\) is an odd function such that 0 is in the domain of \(f\), then \(f(0)=0\).

Short Answer

Expert verified
If \(f\) is an odd function, it satisfies the condition \(f(-x) = -f(x)\) for all values of x in its domain. For x=0, this gives us \(f(0) = -f(0)\). Solving this equation, we get \(f(0) = 0\). Hence, if \(f\) is an odd function with 0 in its domain, \(f(0) = 0\).

Step by step solution

01

Recall the definition of an odd function

An odd function is a function that satisfies the condition: \(f(-x) = -f(x)\) for all values of x in its domain.
02

Evaluate the function at x = 0

We need to show that f(0) = 0. So let's plug x = 0 into the condition for an odd function: \[f(-0) = -f(0)\] Since -0 is the same as 0, this simplifies to: \[f(0) = -f(0)\]
03

Solve the equation for f(0)

We have the equation: \[f(0) = -f(0)\] To solve for f(0), add f(0) to both sides of the equation: \[f(0) + f(0) = 0\] This simplifies to: \[2f(0) = 0\]
04

Divide by 2 to find the value of f(0)

Divide both sides of the equation by 2 to find the value of f(0): \[\frac{2f(0)}{2} = \frac{0}{2}\] This simplifies to: \[f(0) = 0\] We have shown that for an odd function with 0 in its domain, \(f(0) = 0\).

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Key Concepts

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

Function Properties
Understanding function properties aids in analyzing and classifying functions according to specific criteria.
One important property is whether a function is odd or even.
An **odd function** is defined by the relationship \( f(-x) = -f(x) \) for all \( x \) in its domain.In simple terms, if you reflect an odd function across the origin, it looks the same.
This means that the graph of an odd function is symmetric about the origin.
Recognizing these properties makes it easier to predict the function's behavior without intensive graphing or analysis.
Domain of a Function
The domain of a function refers to all the possible input values \( x \) for which the function is defined.
For the function to be considered odd, the domain must include both \( x \) and \( -x \), ensuring the definition \( f(-x) = -f(x) \) holds true.In this specific exercise, the key point was the inclusion of 0 in the function's domain.
Ensuring 0 is within the domain of an odd function allows us to evaluate \( f(0) \) and establish its necessary condition as we did in the solution.
Always check the domain first when working with functions to avoid errors during evaluations.
Function Evaluation
Evaluating a function involves substituting a specific value into a function to see the output.
This process checks how the function behaves at individual points.For odd functions, substituting a point and its negative provides insights into its symmetrical properties.
In our problem, evaluating at \( x = 0 \) shows unique behavior because ordinarily, an odd function is symmetric around zero.
By setting \( f(-0) = -f(0) \), we explore this for the specific case \( x = 0 \).
Such evaluations are fundamental in functioning analysis and solving equations related to functions.
Solving Equations
Solving equations is a key skill in analyzing functions, especially when trying to find specific values like \( f(0) \).
In the given example, solving the equation \( f(0) = -f(0) \) helps demonstrate that \( f(0) \) must indeed equal 0.This is achieved by manipulating the equation
  • Add \( f(0) \) to both sides: \( f(0) + f(0) = 0 \)
  • Simplify to: \( 2f(0) = 0 \)
  • Divide by 2: \( f(0) = 0 \)
These steps solidify understanding of algebraic manipulation and how it can prove properties of functions.
Familiarity with these processes enhances problem-solving skills for various mathematical problems.

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Most popular questions from this chapter

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.

Give an example of a one-to-one function whose domain equals the set of integers and whose range equals the set of positive integers.

For each of the functions \(f\) given in Exercises \(13-\) 22: (a) Find the domain of \(f\). (b) Find the range of \(f\). (c) Find a formula for \(f^{-1}\). (d) Find the domain of \(\boldsymbol{f}^{-1}\). (e) Find the range of \(f^{-1}\). You can check your solutions to part (c) by verify. ing that \(f^{-1} \circ f=I\) and \(f \circ f^{-1}=I\) (recall that \(I\) is the function defined by \(I(x)=x\). $$ f(x)=\frac{1}{3 x+2} $$

Give an example of a function \(f\) such that the domain of \(f\) and the range of \(f\) both equal the set of integers, but \(f\) is not a one-to-one function.

Give an example of two increasing functions whose product is not increasing. [Hint: There are no such examples where both functions are positive everywhere.]
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