/*! 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 17 Show that if \(f^{\prime \prime}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 17

Show that if \(f^{\prime \prime}>0\) throughout an interval \([a, b],\) then \(f^{\prime}\) has at most one zero in \([a, b] .\) What if \(f^{\prime \prime}<0\) throughout \([a, b]\) instead?

Short Answer

Expert verified
\(f'\) has at most one zero for both \(f''>0\) and \(f''<0\) in \([a, b]\).

Step by step solution

01

Understanding the Problem

We need to show that if the second derivative of a function \(f\), \(f''(x)\), is greater than zero on the interval \([a, b]\), then the first derivative \(f'(x)\) can have at most one zero in that interval. Additionally, we will consider the case where \(f''(x) < 0\) and explore its implications.
02

Analyzing the Case: \(f''(x) > 0\)

If \(f''(x) > 0\) on \([a, b]\), \(f'(x)\) is strictly increasing throughout that interval. This means that once \(f'(x)\) goes from negative to positive (or crosses zero), it cannot return to zero again because it keeps increasing. Therefore, \(f'(x)\) can have at most one zero in \([a, b]\).
03

Analyzing the Case: \(f''(x) < 0\)

If \(f''(x) < 0\) on \([a, b]\), then \(f'(x)\) is strictly decreasing throughout that interval. This implies that \(f'(x)\) can again have at most one zero in the interval because once it crosses zero (going from positive to negative), it will continue to decrease and cannot return to zero.
04

Conclusion for Both Cases

For \(f''(x) > 0\), \(f'(x)\) is strictly increasing, allowing at most one zero. For \(f''(x) < 0\), \(f'(x)\) is strictly decreasing, similarly allowing at most one zero. In both cases, the sign of \(f''(x)\) ensures \(f'(x)\) does not change direction more than once in the interval, limiting the number of zeros to at most 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.

First Derivative
The first derivative of a function, denoted as \(f'(x)\), is a vital tool in calculus used to determine the behavior of a function. Specifically, it tells us how the function is changing at any given point. This change can be interpreted in terms of slope: a positive \(f'(x)\) indicates that the function is rising, while a negative \(f'(x)\) suggests it is falling. A zero value means the function has a horizontal tangent at that point, often indicating a potential maximum, minimum, or inflection point.

When dealing with the derivative, the concept of tangents becomes clear. Imagine moving along a curve; the first derivative tells you the slope of the tangent line at each point. Therefore, understanding the first derivative is crucial for graph analysis and optimization problems. The behavior of \(f'(x)\)鈥攚hether it is increasing or decreasing鈥攔eveals more about the nature of the function itself.
Increasing and Decreasing Functions
Functions are classified as increasing or decreasing based on the behavior of their first derivative. When \(f'(x) > 0\) for every \(x\) in a particular interval, the function is said to be increasing on that interval. Conversely, when \(f'(x) < 0\), the function is decreasing. These intervals offer insights into the overall trend and direction of the function.

Understanding this is crucial in analyzing the shape of the graph of \(f(x)\). For example, if the first derivative changes from positive to negative, the function transitions from increasing to decreasing, indicating a peak or local maximum. Conversely, a change from negative to positive shows a trough or local minimum.
  • Strictly Increasing: Function values increase as you move along the interval, proved by positive first derivatives.
  • Strictly Decreasing: Function values constantly decrease, characterized by negative first derivatives.
Recognizing these patterns can immensely help in predicting the behavior of \(f(x)\) without needing to graph it extensively.
Zeros of a Function
Zeros of a function \(f(x)\), or roots, are simply the values of \(x\) for which \(f(x) = 0\). These zeros are significant as they represent the intercepts on the x-axis of a graph. Finding these points is essential in solving equations and understanding the function's behavior on its domain.

Besides the main function, the zeros of the first derivative \(f'(x)\) are equally critical in analysis. These points often indicate where the function has reached a peak or valley, marking critical points where the growth or decline of the function changes. It is essential to analyze these zeros within their interval because they can signify turning points, giving insight into the function's graph and facilitating the understanding of its concavity and overall shape.
  • Identifying Zeros: Usually involves setting \(f(x)\) or \(f'(x)\) to zero and solving for \(x\).
  • Significance: Provides crucial information about intercepts, peaks, and valleys.
Effectively locating zeros allows for deeper insights into the behavior and intricacies of the function.

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

Stopping a car in time You are driving along a highway at a steady \(108 \mathrm{km} / \mathrm{h}(30 \mathrm{m} / \mathrm{s})\) when you see an accident ahead and slam on the brakes. What constant deceleration is required to stop your car in \(75 \mathrm{m} ?\) To find out, carry out the following steps. 1. Solve the initial value problem Differential equation: \(\frac{d^{2} s}{d t^{2}}=-k\) \((k \text { constant })\) Initial conditions: \(\quad \frac{d s}{d t}=30\) and \(s=0\) when \(t=0\) 2\. Find the value of \(t\) that makes \(d s / d t=0 .\) (The answer will involve \(k .)\) 3\. Find the value of \(k\) that makes \(s=75\) for the value of \(t\) you found in Step 2.

Motion with constant acceleration The standard equation for the position \(s\) of a body moving with a constant acceleration \(a\) along a coordinate line is $$where \(v_{0}\) and \(s_{0}\) are the body's velocity and position at time \(t=0 .\) Derive this equation by solving the initial value problem Differential equation: \(\frac{d^{2} s}{d t^{2}}=a\) Initial conditions: \(\quad \frac{d s}{d t}=v_{0}\) and \(s=s_{0}\) when \(t=0\)s=\frac{a}{2} t^{2}+v_{0} t+s_{0}$$ where \(v_{0}\) and \(s_{0}\) are the body's velocity and position at time \(t=0 .\) Derive this equation by solving the initial value problem (1).

Right, or wrong? Give a brief reason why. $$\int \frac{-15(x+3)^{2}}{(x-2)^{4}} d x=\left(\frac{x+3}{x-2}\right)^{3}+C$$

One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is $$A(q)=\frac{k m}{q}+c m+\frac{h q}{2}$$ where \(q\) is the quantity you order when things run low (shoes, radios, brooms, or whatever the item might be), \(k\) is the cost of placing an order (the same, no matter how often you order), \(c\) is the cost of one item (a constant), \(m\) is the number of items sold each week (a constant), and \(h\) is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security). a. Your job, as the inventory manager for your store, is to find the quantity that will minimize \(A(q) .\) What is it? (The formula you get for the answer is called the Wilson lot size formula.) b. Shipping costs sometimes depend on order size. When they do, it is more realistic to replace \(k\) by \(k+b q,\) the sum of \(k\) and a constant multiple of \(q\). What is the most economical quantity to order now?

a. The function \(y=\cot x-\sqrt{2} \csc x\) has an absolute maximum value on the interval \(0
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

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.