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Chapter 1: Problem 55

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

Short Answer

Expert verified
True. If \(f\) is an even function whose domain is the set of real numbers, and a function \(g\) is defined by the given piecewise function, then \(g\) is an odd function. This is because \(g\) satisfies the condition for an odd function: \(g(-x) = -g(x), \forall x \in \mathbb{R}\).

Step by step solution

01

Recall the properties of even and odd functions

An even function satisfies \(f(-x) = f(x), \forall x \in \mathbb{R}\). An odd function satisfies \(f(-x) = -f(x), \forall x \in \mathbb{R}\).
02

Analyze the function \(g\)

The function \(g(x)\) is defined as: $$ g(x)=\left\\{\begin{array}{ll} f(x) & \text { if } x \geq 0 \\\ -f(x) & \text { if } x<0 \end{array}\right. $$ We need to check if \(g\) satisfies the condition for an odd function: \(g(-x) = -g(x), \forall x \in \mathbb{R}\).
03

Check the condition for positive x and negative x separately

First, let's check the condition for \(x \geq 0\). If \(x \geq 0\), then \(-x \leq 0\), so \(g(-x) = -f(-x)\). In this case, we have: $$ g(-x) = -f(-x) = -f(x) = -g(x) $$ where we used that \(f(-x) = f(x)\) since \(f\) is even. Now, let's check the condition for \(x < 0\). If \(x < 0\), then \(-x > 0\), so \(g(-x) = f(-x)\). In this case, we have: $$ g(-x) = f(-x) = f(x) = -(-f(x)) = -g(x) $$ where we again used that \(f(-x) = f(x)\) since \(f\) is even.
04

Conclude if \(g\) is odd or not

Since both conditions are satisfied, i.e., \(g(-x) = -g(x)\) for both \(x \geq 0\) and \(x < 0\), we can conclude that \(g\) is an odd function. So, the given statement is true: If \(f\) is an even function whose domain is the set of real numbers, and a function \(g\) is defined by the given piecewise function, then \(g\) is an odd function.

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Most popular questions from this chapter

Suppose \(f\) is the function whose domain is the interval \([-2,2],\) with \(f\) defined by the following formula: $$ f(x)=\left\\{\begin{array}{ll} -\frac{x}{3} & \text { if }-2 \leq x<0 \\ 2 x & \text { if } 0 \leq x \leq 2 \end{array}\right. $$ (a) Sketch the graph of \(f\). (b) Explain why the graph of \(f\) shows that \(f\) is not a one-to-one function. (c) Give an explicit example of two distinct numbers \(a\) and \(b\) such that \(f(a)=f(b)\).

(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.

A constant function is a function whose value is the same at every number in its domain. For example, the function \(f\) defined by \(f(x)=4\) for every number \(x\) is a constant function. Give an example of three functions \(f, g,\) and \(h\), none of which is a constant function, such that \(f \circ h=g \circ h\) but \(f\) is not equal to \(g\).

For Exercises \(33-40,\) assume that \(f\) is the function defined by $$ f(x)=\left\\{\begin{array}{ll} 2 x+9 & \text { if } x<0 \\ 3 x-10 & \text { if } x \geq 0 \end{array}\right. $$ Evaluate \(f(1)\).

Give an example of a function whose domain equals (0,1) and whose range equals [0,1] .
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