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Chapter 1: Problem 39

Give an example to show that the product of two one-to-one functions is not necessarily a one-to-one function.

Short Answer

Expert verified
Let \(f(x) = x\) and \(g(x) = x + 1\), which are both one-to-one functions. Their product function is \(h(x) = f(x) \cdot g(x) = x(x+1)\). Evaluating \(h(x)\) for \(x=-1\) and \(x=0\) gives \(h(-1) = h(0) = 0\), even though \(-1 \neq 0\). This demonstrates that the product of \(f(x)\) and \(g(x)\) is not a one-to-one function, as there are different values of \(x\) which result in the same value of \(h(x)\).

Step by step solution

01

Define two one-to-one functions

Let's define two simple one-to-one functions: Function \(f(x)\): \( f(x) = x \) Function \(g(x)\): \( g(x) = x+1 \) These two functions are one-to-one because if \(x_1 \neq x_2 \), then \(f(x_1) \neq f(x_2)\) and \(g(x_1) \neq g(x_2)\).
02

Find the product of the two functions

Now we need to find the product of these two functions: \(h(x) = f(x) \cdot g(x) = x(x+1)\)
03

Show that the product is not one-to-one

In order to show that the product function \(h(x)\) is not one-to-one, we need to find at least two different values of \(x\) for which \(h(x)\) is the same. Let's test the function for \(x=-1\) and \(x=0\). For \(x=-1\): \(h(-1)=(-1)(0) = 0\) For \(x=0\): \(h(0)=(0)(1) = 0\) We can see that \(h(-1) = h(0)\), even though \(-1 \neq 0\). This demonstrates that the product of these two one-to-one functions is not one-to-one, as there are different values of \(x\) which result in the same value of \(h(x)\).

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Most popular questions from this chapter

Suppose \(f\) is the function whose domain is the set of real numbers, with \(f\) defined on this domain by the formula $$ f(x)=|x+6| $$ Explain why \(f\) is not a one-to-one function.

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.

Assume that \(f\) is the function defined by $$ f(x)=\left\\{\begin{array}{ll} 2 x+9 & \text { if } x<0 \\ 3 x-10 & \text { if } x \geq 0. \end{array}\right. $$ Find two different values of \(x\) such that \(f(x)=0\).

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 \(f(-2)\).

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 odd function. Evaluate \(f(-2)\).
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