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True or false: If \(f\) is an odd function whose domain is the set of real numbers and a function \(g\) is defined by $$ g(x)=\left\\{\begin{array}{ll} f(x) & \text { if } x \geq 0 \\ -f(x) & \text { if } x<0 \end{array}\right. $$

Short Answer

Expert verified
The statement is false. Although the function \(f\) is odd, the function \(g\) defined by the given piecewise function does not satisfy the property of odd functions (\(g(-x) = -g(x)\)) for all values of \(x\), specifically when \(x \geq 0\).

Step by step solution

01

Define an odd function

Recall the definition of an odd function. A function \(f\) is considered odd if for all \(x\) in the domain of the function, the following condition holds: \[f(-x) = -f(x).\]
02

Analyze the properties of the function g

The function \(g\) is defined piecewise as follows: \[ g(x)=\left\\{\begin{array}{ll} f(x) & \text { if } x \geq 0 \\\ -f(x) & \text { if } x<0 \end{array}\right. \] We can analyze how the definition of \(g\) is related to the properties of the function \(f\) by analyzing the behavior of \(g(x)\) for both positive and negative values of \(x\), and comparing them to the properties of odd functions.
03

Analyze the properties of the function g for x ≥ 0

If \(x \geq 0\), then the definition of \(g(x)\) is given by: \[g(x) = f(x).\] Now, let's find the value of \(g(-x)\) for \(x \geq 0\). Since \(x \geq 0\), we have that \(-x \leq 0\). Thus, the definition of \(g(-x)\) in this case is given by: \[g(-x) = -f(-x).\] Since function \(f\) is odd, we have \(f(-x) = -f(x)\). Therefore, we can substitute this back into the expression for \(g(-x)\): \[g(-x) = -(-f(x))\Rightarrow g(-x) = f(x).\] So, for \(x \geq 0\), we have that \(g(-x) = g(x)\). Notice this is different from the requirement for being an odd function: \(g(-x) = -g(x)\).
04

Draw the conclusion

We found that for \(x \geq 0\), the relationship between \(g(-x)\) and \(g(x)\) is given by \(g(-x) = g(x)\), which is different from the requirement for being an odd function (\(g(-x) = -g(x)\)). Therefore, the given statement is false: if \(f\) is an odd function whose domain is the set of real numbers and a function \(g\) is defined by the given piecewise function, then the function \(g\) is not an odd function.

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Key Concepts

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

Understanding Piecewise Functions
A piecewise function is defined by different expressions for different intervals of its domain. This is much like a chameleon changing its colors to adapt to different environments. To put it simply, we have a recipe with different instructions depending on the ingredient — in this case, the input value.

When looking at our textbook problem, we see that the function g(x) changes its formula based on whether x is non-negative or negative. If x ≥ 0, we stick to the original function f(x), much like following the same route home. If x < 0, we flip the sign of f(x), which can be viewed as taking a detour on our way back. Understanding piecewise functions is crucial because they can model real-world scenarios where behaviour changes under different conditions.
Delving into Function Analysis
The heart of function analysis lies in exploring the characteristics and behaviours of a function. Whenever we face a new function, think of it as meeting a new person. We want to get to know them — how they look, how they behave in situations, and what their defining traits are.

In function analysis, we dissect functions piece by piece. We look at intervals, identify odd or even properties, pin down where they increase or decrease, and scrutinize how they behave near critical points. Let's take our function g(x) from the exercise. It's crucial to study its behavior for both positive and negative x values. The analysis revealed that g(x) mirrors f(x) when x is non-negative but opposes it when x is negative. This subtle nuance is pivotal in determining the overall nature of the function — and whether it adheres to being odd or even.
The Prelude of Precalculus
In the grand storyline of mathematics, precalculus is the prelude to the adventure of calculus. It sets the stage with crucial concepts like functions, equations, and their properties. Precalculus is where we tie the lace of our mathematical boots, preparing to embark on the hike through the landscapes of limits, derivatives, and integrals.

It's in precalculus that we learn about the idea of odd and even functions — a pair of core types each with its mirrored charm. An odd function, like our f(x) in the exercise, is symmetrical around the origin, meaning it appears the same right side up or upside down. This is different from even functions, which are symmetrical about the y-axis, like a perfect reflection in a mirror. Knowing these types of functions is essential because they foretell the behavior of more complicated functions encountered in calculus.

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Most popular questions from this chapter

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 \(I(x)=x\). $$ \begin{aligned} &f(x)=x^{2}+8, \text { where the domain of } f \text { equals }\\\ &(0, \infty) \end{aligned} $$

(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.

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

Suppose \(f\) and \(g\) are functions, each of whose domain consists 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 \(f^{-1} \circ f\).

(a) True or false: The product 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 product 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.

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