/*! 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 27 Use the MVT to prove that $$ ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 27

Use the MVT to prove that $$ \frac{1}{3(m+1)^{2 / 3}}<(m+1)^{1 / 3}-m^{1 / 3}<\frac{1}{3 m^{2 / 3}} \quad \text { for all } m \in \mathbb{N} $$

Short Answer

Expert verified
Using the Mean Value Theorem (MVT) on the function \(f(x) = x^{1/3}\) for the interval \([m, m+1]\), we find a number \(c \in (m, m+1)\) such that \((m+1)^{1/3} - m^{1/3} = \frac{1}{3} c^{-2/3}\). Since \(m < c < m+1\), we have \(\frac{1}{(m+1)^{2/3}} < c^{-2/3} < \frac{1}{m^{2/3}}\). Multiplying by \(\frac{1}{3}\), we obtain the desired inequality: \(\frac{1}{3(m+1)^{2/3}} < (m+1)^{1/3}-m^{1/3} < \frac{1}{3m^{2/3}}\).

Step by step solution

01

Consider the function \(f(x) = x^{1/3}\) on the interval \([m, m+1]\), with \(m \in \mathbb{N}\). Note that \(f(x)\) is continuous and differentiable on this interval since it is a power function. #Step 2: Apply the Mean Value Theorem#

By applying the MVT to the function \(f(x) = x^{1/3}\) on the interval \([m, m+1]\), we obtain that there exists a number \(c \in (m, m+1)\) such that \[ \frac{f(m+1)-f(m)}{(m+1)-m} = f'(c) \] #Step 3: Calculate the derivative#
02

Now, let's find the derivative of \(f(x)\): \[ f'(x) = \frac{d}{dx} x^{1/3} = \frac{1}{3} x^{-2/3} \] #Step 4: Substitute the derivative and simplify#

Substituting the derivative into our MVT equation, we get \[ \frac{(m+1)^{1/3}-m^{1/3}}{1}= \frac{1}{3} c^{-2/3} \] which implies that \[ (m+1)^{1/3} - m^{1/3} = \frac{1}{3} c^{-2/3} \] #Step 5: Bound the function with the given inequality#
03

Here, we have \(m < c < m+1\). Therefore, we can derive the following bounds for \(c^{-2/3}\): \[ \frac{1}{(m+1)^{2/3}} < c^{-2/3} < \frac{1}{m^{2/3}} \] #Step 6: Multiply both sides by 1/3#

To obtain the desired inequality, we multiply both sides of the above inequality by 1/3: \[ \frac{1}{3(m+1)^{2/3}} < \frac{1}{3}c^{-2/3} < \frac{1}{3m^{2/3}} \] which can be rewritten as \[ \frac{1}{3(m+1)^{2/3}} < (m+1)^{1/3}-m^{1/3} < \frac{1}{3m^{2/3}} \] Thus, we have proven the given inequality using the Mean Value Theorem.

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!

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

Use the MVT to prove that for all \(n \in \mathbb{N}\) and \(a, b \in \mathbb{R}\) such that \(0

Given \(n \in \mathbb{N}\), let \(f_{n}: \mathbb{R} \rightarrow \mathbb{R}\) be defined by \(f_{n}(x):=x^{n}\) if \(x \geq 0\) and \(f_{n}(x):=-x^{n}\) if \(x<0 .\) Show that \(f_{n}\) is \((n-1)\) -times differentiable on \(\mathbb{R}\), \(f_{n}^{(n-1)}\) is continuous on \(\mathbb{R}\), but \(f_{n}^{(n)}(0)\) does not exist.

Show that the \(x\) -axis is a normal to the curve \(y^{2}=x\) at \((0,0)\). If three normals can be drawn to this curve from a point \((a, 0)\), show that \(a\) must be greater than \(\frac{1}{2}\). Find the value of \(a\) such that the two normals, other than the \(x\) -axis, are perpendicular to each other.

Show that the cubic \(2 x^{3}+3 x^{2}+6 x+10\) has exactly one real root.

Suppose \(f, g: \mathbb{R} \rightarrow \mathbb{R}\) satisfy \(f(x-y)=f(x) g(y)-g(x) f(y)\) and \(g(x-y)=g(x) g(y)+f(x) f(y)\) for all \(x, y \in \mathbb{R}\). If \(f_{+}^{\prime}(0)\) exists, then show that \(f\) and \(g\) are differentiable at every \(c \in \mathbb{R}\), and \(f^{\prime}(c)=f^{\prime}(0) g(c)\) and \(g^{\prime}(c)=-f^{\prime}(0) f(c) .\) In fact, show that \(f\) and \(g\) are infinitely differentiable. If \(f_{+}^{\prime}(0)=2\), find \(f^{(n)}(1)\) and \(g^{(n)}\) (1) in terms of \(f(1)\) and \(g(1)\). (Hint: Prove that \(f\) is an odd function, \(g\) is an even function, \(f\) and \(g\) are differentiable at 0 and \(g^{\prime}(0)=0 .\) Use Exercise 6.)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

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.