/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 Monotonic Functions Show that mo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 63

Monotonic Functions Show that monotonic increasing and decreasing functions are one-to-one.

Short Answer

Expert verified
A monotonic function is one-to-one if for any two different points in its domain, no two have the same image. This is intrinsic to the property of being monotonically increasing or decreasing.

Step by step solution

01

Case 1 Monotonically Increasing

Assume a function \( f(x) \) is monotonically increasing in its domain. If \( x_1 \) and \( x_2 \) are two different points in the domain and \( x_1 > x_2 \) then by the properties of a monotonically increasing function \( f(x_1) > f(x_2) \). Hence, no two different points in the domain have the same image and the function \( f(x) \) is one-to-one.
02

Case 2 Monotonically Decreasing

Assume a function \( f(x) \) is monotonically decreasing in its domain. If \( x_1 \) and \( x_2 \) are two different points in the domain and \( x_1 > x_2 \) then by the properties of a monotonically decreasing function \( f(x_1) < f(x_2) \). Hence, no two different points in the domain have the same image and the function \( f(x) \) is one-to-one.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

One-to-One Functions
Imagine you are at a party, and every guest is given a unique gift. Just like these gifts, in mathematics, a function is called a one-to-one function (or injective) if it gives a different output for every different input. It's like having a special rule that ensures no one at the party gets the same gift.

For every unique value you put into this function, you're guaranteed to get a unique result back out. Why does this matter? Such functions make it possible to talk about 'reversing' the process to get back the original input, leading to the concept of 'inverses' in math.
Monotonically Increasing Functions
Let's go on a hike, but on a path that only goes up, never down. This is similar to how a monotonically increasing function behaves: for any two different points you choose along its journey, the later one is always higher or at least equal in elevation compared to the earlier one. Mathematically speaking, if you have two inputs, and the second is larger, the output of the second will be greater than or equal to that of the first.

This consistent upward trend guarantees that each input has a distinct output, leading back to our concept of one-to-one functions.
Monotonically Decreasing Functions
Imagine now your hike is on a slope heading strictly downwards. This captures the essence of a monotonically decreasing function: as you move along with the inputs, the outputs fall or remain constant. If you select any two different points where the second is past the first, the result for the second point will always be less than or equal to that of the first.

This never-rising characteristic is what secures the function's status as one-to-one. No matter how many steps you take downhill, you will never end up at a higher point than where you started.
Function Properties
So, what makes a function special? Functions come with a toolkit of properties that describe their behaviors, like the directions they move (increasing or decreasing) and how they pair inputs to outputs (one-to-one).

When a function is both one-to-one and either monotonically increasing or decreasing, we have a clear understanding of how it behaves across its entire domain. These properties help us to analyze the functions, predict outcomes, and even find inverses. They also ensure the function passes what's known as the 'Horizontal Line Test'鈥攁 useful visual aid in determining if a function is one-to-one simply by drawing horizontal lines through its graph.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Multiple Choice If the volume of a cube is increasing at 24 \(\mathrm{in}^{3} / \mathrm{min}\) and the surface area of the cube is increasing at 12 \(\mathrm{in}^{2} / \mathrm{min}\) , what is the length of each edge of the cube? \(\mathrm{}\) \(\begin{array}{lll}{\text { (A) } 2 \text { in. }} & {\text { (B) } 2 \sqrt{2} \text { in. (C) } \sqrt[3]{12} \text { in. (D) } 4 \text { in. }}\end{array}\)

The domain of f^{\prime}\( is \)[0,1) \cup(1,2) \cup(2,3]

Unique Solution Assume that \(f\) is continuous on \([a, b]\) and differentiable on \((a, b) .\) Also assume that \(f(a)\) and \(f(b)\) have op- posite signs and \(f^{\prime} \neq 0\) between \(a\) and \(b\) . Show that \(f(x)=0\) exactly once between \(a\) and \(b .\)

$$ \begin{array}{l}{\text { Multiple Choice Which of the following functions is an }} \\ {\text { antiderivative of } \frac{1}{\sqrt{x}} ? \quad \mathrm{}}\end{array} $$ $$ (\mathbf{A})-\frac{1}{\sqrt{2 x^{3}}}(\mathbf{B})-\frac{2}{\sqrt{x}} \quad(\mathbf{C}) \frac{\sqrt{x}}{2}(\mathbf{D}) \sqrt{x}+5(\mathbf{E}) 2 \sqrt{x}-10 $$

The Effect of Flight Maneuvers on the Heart The amount of work done by the heart's main pumping chamber, the left ventricle, is given by the equation $$W=P V+\frac{V \delta v^{2}}{2 g}$$ where \(W\) is the work per unit time, \(P\) is the average blood pressure, \(V\) is the volume of blood pumped out during the unit of time, \(\delta("\) delta") is the density of the blood, \(v\) is the average velocity of the exiting blood, and \(g\) is the acceleration of gravity. When \(P, V, \delta,\) and \(v\) remain constant, \(W\) becomes a function of \(g,\) and the equation takes the simplified form $$W=a+\frac{b}{g}(a, b\( constant \))$$ As a member of NASA's medical team, you want to know how sensitive \(W\) is to apparent changes in \(g\) caused by flight maneuvers, and this depends on the initial value of \(g\) . As part of your investigation, you decide to compare the effect on \(W\) of a given change \(d g\) on the moon, where \(g=5.2 \mathrm{ft} / \mathrm{sec}^{2},\) with the effect the same change \(d g\) would have on Earth, where \(g=32\) \(\mathrm{ft} / \mathrm{sec}^{2} .\) Use the simplified equation above to find the ratio of \(d W_{\mathrm{moon}}\) to \(d W_{\mathrm{Earth}}\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.