Problem 31 Let the real function $f$ be s... [FREE SOLUTION]

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Transition to Higher Mathematics: Structure and Proof

Bob A. Dumas, John E. McCarthy

$Math Studyset 91影视 Explanations$ Math

2 Edition

Chapter 1: Problem 31

Let the real function $f$ be strictly increasing. Show that for any $b \in \mathbb{R}, f^{-1}(b)$ is either empty or consists of a single element, and that $f$ is therefore an injection. If $f$ is also a bijection, is the inverse function of $f$ also strictly increasing?

Short Answer

Expert verified

$f^{-1}(b)$ is either empty or single-element; $f$ is injective. If bijective, $f^{-1}$ is also strictly increasing.

Step by step solution

Understanding the Properties of a Strictly Increasing Function

A function $f$ is strictly increasing if for any two numbers $x_1$ and $x_2$ in its domain, $x_1 < x_2$ implies $f(x_1) < f(x_2)$. This property means the function never decreases or remains constant.

Prove that $f^{-1}(b)$ is Empty or a Single Element

Assume $f(x_1) = b = f(x_2)$ and $x_1 eq x_2$. Since $f$ is strictly increasing, $x_1 < x_2$ implies $f(x_1) < f(x_2)$, which is a contradiction. Thus, if $f(x) = b$, then $x$ is unique, meaning $f^{-1}(b)$ is either empty or contains exactly one element.

Show that $f$ is an Injection

By the definition of injectivity, a function $f$ is injective if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Our previous proof that $f(x_1) = f(x_2)$ leads to $x_1 = x_2$ shows that $f$ is injective.

Consideration if $f$ is Also a Bijection

If $f$ is bijective, it has an inverse function $f^{-1}$. To check if $f^{-1}$ is strictly increasing, take any $b_1 < b_2$. Since $f$ is strictly increasing and surjective (bijective), $b_1 = f(x_1)$ and $b_2 = f(x_2)$ implies $x_1 < x_2$, suggesting $f^{-1}(b_1) < f^{-1}(b_2)$. Therefore, $f^{-1}$ is strictly increasing.

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Key Concepts

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

Injective Function

An injective function, also known as a one-to-one function, is a function where every element in the codomain is mapped by at most one element from the domain. In simpler terms, no two different input values map to the same output value. This property can be effectively used to understand whether a function is injective:

If for every pair of distinct inputs, say $x_1$ and $x_2$, the outputs $f(x_1)$ and $f(x_2)$ are different, then the function is injective.
If a function is represented graphically, an injective function will not have any horizontal line intersect the graph more than once.

In our exercise, the strictly increasing nature of the function ensures that if $f(x_1) = f(x_2)$, it must be true that $x_1 = x_2$. This confirms our function is injective.

Inverse Function

An inverse function essentially reverses the effect of the original function. For a given function $f$, the inverse $f^{-1}$ maps items from the codomain back to the domain, restoring the original input used in $f$.

For a function to have an inverse, two main conditions must be satisfied:

The function must be injective, meaning it passes the horizontal line test explained above.
The function must be surjective, meaning every element in the codomain is the image of an element from the domain.

Since a strictly increasing function is injective, it can have an inverse. In the context of this exercise, if $f$ is strictly increasing and covers its range completely (bijective), then it is invertible, and $f^{-1}$ itself will also be strictly increasing. Each element $b$ then maps back to exactly one element from the domain, emphasizing the uniqueness of the inverse.

Bijection

A bijection is a special type of function that is both injective and surjective. This means every element of the codomain is assigned to exactly one unique element of the domain, and no elements in the codomain are left unassigned. Hence, bijective functions are sometimes referred to as "one-to-one and onto" functions.

Injective (one-to-one): The function doesn't map distinct elements in the domain to the same element in the codomain.
Surjective (onto): Every element of the codomain is the image of at least one element from the domain.

In a mathematical context, bijections are ideal for ensuring that an inverse function can be defined. In our initial problem, if $f$ is strictly increasing (injective) and bijective, it implies every element of the real numbers is covered, and thus it possesses a valid inverse function. This inverse will also be strictly increasing due to the nature of bijections.

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91影视

Let the real function \(f\) be strictly increasing. Show that for any \(b \in \mathbb{R}, f^{-1}(b)\) is either empty or consists of a single element, and that \(f\) is therefore an injection. If \(f\) is also a bijection, is the inverse function of \(f\) also strictly increasing?

Short Answer

Step by step solution

Understanding the Properties of a Strictly Increasing Function

Prove that \(f^{-1}(b)\) is Empty or a Single Element

Show that \(f\) is an Injection

Consideration if \(f\) is Also a Bijection

Key Concepts

Injective Function

Inverse Function

Bijection

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