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Chapter 1: Problem 66

Give an example of two different functions \(f\) and \(g\), both of which have the set of real numbers as their domain, such that \(f(x)=g(x)\) for every rational number \(x\).

Short Answer

Expert verified
The two functions that satisfy the given condition are: 1. \(f(x) = x\) 2. \(g(x) = \begin{cases} x & \text{if } x \text{ is rational} \\ -x & \text{if } x \text{ is irrational} \\ \end{cases} \) These functions have the same values for every rational number x, and different values for irrational numbers.

Step by step solution

01

Choose function f

We can pick a simple function for f. Let's use the identity function: \(f(x) = x\).
02

Define function g

We need to find another function, g, that has the same values as f for every rational number, meaning \(g(x)=x\) for every rational number x, but is still different from f. Consider the piecewise function g: \[g(x) = \begin{cases} x & \text{if } x \text{ is rational} \\ -x & \text{if } x \text{ is irrational} \\ \end{cases} \]
03

Comparing functions f and g

Now, let's compare these two functions, f and g, to verify if they satisfy the condition given: 1. For any rational number x, f(x) = x and g(x) = x, so f(x) = g(x) for every rational number x, which is required according to the exercice. 2. For irrational number x, f(x) = x and g(x) = -x. We can see that the functions are different for irrational numbers. Thus, we have found two functions that satisfy the given condition: f(x) = x and g(x) is a piecewise function defined by the above rules.

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Most popular questions from this chapter

Suppose you are exchanging cur. rency in the London airport. The currency exchange service there only makes transactions in which one of the two currencies is British pounds, but you want to exchange dollars for Euros. Thus you first need to exchange dollars for British pounds, then exchange British pounds for Euros. At the time you want to make the exchange, the function \(f\) for exchanging dollars for British pounds is given by the formula \(f(d)=0.66 d-1\) and the function \(g\) for exchanging British pounds for Euros is given by the formula \(g(p)=1.23 p-2\) The subtraction of 1 or 2 in the number of British pounds or Euros that you receive is the fee charged by the currency exchange service for each transaction. Which process gives you more Euros: exchanging $$\$ 100$$ for Euros twice or exchanging $$\$ 200$$ for Euros once?

Suppose \(f\) and \(g\) are functions, each with domain of four numbers, with \(f\) and \(g\) defined by the tables below: $$\begin{array}{c|c}x & f(x) \\\\\hline 1 & 4 \\\2 & 5 \\\3 & 2 \\\4 & 3\end{array}$$ $$\begin{array}{c|c}x & g(x) \\\\\hline 2 & 3 \\\3 & 2 \\\4 & 4 \\\5 & 1\end{array}$$ Sketch the graph of \(f^{-1}\).

Suppose $$h(x)=\left(\frac{x^{2}+1}{x-1}-1\right)^{3}$$ (a) If \(f(x)=x^{3},\) then find a function \(g\) such that \(h=f \circ g\) (b) If \(f(x)=(x-1)^{3}\), then find a function \(g\) such that \(h=f \circ g\).

For each of the functions \(f\) given. (a) Find the domain of \(f\). (b) Find the range of \(f\). (c) Find a formula for \(f^{-1}\). (d) Find the domain of \(f^{-1}\). (e) Find the range of \(f^{-1}\). You can check your solutions to part (c) by verifying that \(f^{-1} \circ f=I\) and \(f \circ f^{-1}=I\) (recall that \(I\) is the function defined by \(I(x)=x\) ). \(f(x)=2 x^{2}+5,\) where the domain of \(f\) equals \((0, \infty)\).

Suppose \(f\) and \(g\) are functions, each with domain of four numbers, with \(f\) and \(g\) defined by the tables below: $$\begin{array}{c|c}x & f(x) \\\\\hline 1 & 4 \\\2 & 5 \\\3 & 2 \\\4 & 3\end{array}$$ $$\begin{array}{c|c}x & g(x) \\\\\hline 2 & 3 \\\3 & 2 \\\4 & 4 \\\5 & 1\end{array}$$ What is the range of \(g^{-1} ?\)
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