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Are the following functions one-to-one over their domain of existence? Does the function have an inverse? If so, find the inverse \(f^{-1}(x)\) of the function. Justify your answer. $$ f(x)=\frac{1}{x} $$

Short Answer

Expert verified
Yes, \( f(x) = \frac{1}{x} \) is one-to-one and has an inverse, which is \( f^{-1}(x) = \frac{1}{x} \).

Step by step solution

01

Determining if the Function is One-to-One

A function is one-to-one if every distinct input gives a distinct output. For \( f(x) = \frac{1}{x} \), assume \( f(a) = f(b) \). This means \( \frac{1}{a} = \frac{1}{b} \). Therefore, cross-multiplying gives \( a = b \). This indicates that any two inputs result in different outputs unless the inputs are the same, thus \( f(x) = \frac{1}{x} \) is one-to-one.
02

Checking the Domain of the Function

The function \( f(x) = \frac{1}{x} \) is defined for all real numbers except \( x = 0 \). Therefore, its domain is \( x \in \mathbb{R} \setminus \{0\} \). This is important for considering the existence of an inverse.
03

Finding the Inverse of the Function

Since \( f(x) = \frac{1}{x} \) is one-to-one over its domain, it has an inverse. To find the inverse, let \( y = \frac{1}{x} \). Solve for \( x \) in terms of \( y \): Multiply both sides by \( x \) to get \( xy = 1 \). Then solve for \( x \) by setting \( x = \frac{1}{y} \). So, the inverse function \( f^{-1}(x) = \frac{1}{x} \).
04

Justification of the Inverse

The inverse function \( f^{-1}(x) = \frac{1}{x} \) is valid within the same domain as the original function, excluding zero. We can verify that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \) for each valid \( x \), confirming the inverse relationship.

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Key Concepts

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

One-to-One Function
A function is considered one-to-one if each output value is uniquely paired with one input value. This means that different inputs should result in different outputs. It's like having a unique fingerprint for each input. To check if a function is one-to-one, we can use the horizontal line test: if a horizontal line intersects the graph of the function more than once, it's not one-to-one. In our example, the function given is \( f(x) = \frac{1}{x} \). We assumed \( f(a) = f(b) \), leading us to \( \frac{1}{a} = \frac{1}{b} \). By cross-multiplying, we conclude that \( a = b \). This shows that any two distinct inputs yield distinct outputs, confirming that \( f(x) = \frac{1}{x} \) is indeed one-to-one.
  • A one-to-one function ensures each output is paired with only one input.
  • The horizontal line test is a helpful tool for identification.
  • In our case, \( \frac{1}{a} = \frac{1}{b} \) implies \( a = b \), verifying the function is one-to-one.
Domain of a Function
The domain of a function defines all the possible input values that the function can accept. It's essential to understand the domain when dealing with functions because it tells us the limits within which the function operates. For \( f(x) = \frac{1}{x} \), we look for values of \( x \) for which the function is defined. Given that division by zero is undefined, \( x = 0 \) is excluded from the domain. Thus, the domain of \( f(x) = \frac{1}{x} \) is all real numbers except zero, which is mathematically expressed as \( x \in \mathbb{R} \setminus \{0\} \). Recognizing this exclusion is crucial for both the function and its inverse.
  • The domain is the set of all inputs a function can accept.
  • For \( f(x) = \frac{1}{x} \), the input cannot be zero.
  • Thus, the domain is all real numbers except zero: \( \mathbb{R} \setminus \{0\} \).
Finding Inverses
Finding the inverse of a function involves reversing the roles of inputs and outputs. If our function maps \( x \) to \( y \), its inverse will map \( y \) back to \( x \). To find an inverse, we typically solve for the original input variable in terms of the output. For \( f(x) = \frac{1}{x} \), we set it equal to \( y \), giving us \( y = \frac{1}{x} \). Solving for \( x \), by multiplying both sides by \( x \), we find \( xy = 1 \). We then isolate \( x \) as \( x = \frac{1}{y} \). With this, the inverse function, denoted as \( f^{-1}(x) \), is \( \frac{1}{x} \). It's important to verify that both \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \), ensuring that the inverse is correct within the function's domain, which excludes zero.
  • To find a function's inverse, swap inputs and outputs.
  • For \( f(x) = \frac{1}{x} \), solve \( y = \frac{1}{x} \) for \( x \) to get the inverse.
  • The inverse function is \( f^{-1}(x) = \frac{1}{x} \), ensuring valid mappings back to original inputs.

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Most popular questions from this chapter

The concentration of hydrogen ions in a substance is denoted by \(\left[\mathrm{H}^{+}\right],\) measured in moles per liter. The pH of a substance is defined by the logarithmic function \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] .\) This function is used to measure the acidity of a substance. The pH of water is \(7 .\) A substance with a pH less than 7 is an acid, whereas one that has a pH of more than 7 is a base. a. Find the \(\mathrm{pH}\) of the following substances. Round answers to one digit. b. Determine whether the substance is an acid or a base. i. Eggs: \(\left[\mathrm{H}^{+}\right]=1.6 \times 10^{-8} \mathrm{mol} / \mathrm{L}\) ii. Beer: \(\left[\mathrm{H}^{+}\right]=3.16 \times 10^{-3} \mathrm{molL}\) iii. Tomato Juice: \(\left[\mathrm{H}^{+}\right]=7.94 \times 10^{-5} \mathrm{molL}\)

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