/*! 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 3 Find the value or values of \(c\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 3

Find the value or values of \(c\) that satisfy the equation $$\frac{f(b)-f(a)}{b-a}=f^{\prime}(c)$$ in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises \(1-6 .\) $$f(x)=x+\frac{1}{x}, \quad\left[\frac{1}{2}, 2\right]$$

Short Answer

Expert verified
The value of \(c\) is 1.

Step by step solution

01

Understand the Mean Value Theorem (MVT)

The Mean Value Theorem states that for a function \( f(x) \) continuous on \([a, b]\) and differentiable on \((a, b)\), there is at least one \(c\) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
02

Compute \(f(a)\) and \(f(b)\)

For \(f(x) = x + \frac{1}{x}\), calculate \(f(a)\) and \(f(b)\):\[ f\left(\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{\frac{1}{2}} = \frac{1}{2} + 2 = \frac{5}{2} \]\[ f(2) = 2 + \frac{1}{2} = 2.5 \]
03

Calculate the Slope of the Secant Line

Find the slope of the secant line using \(f(b)\) and \(f(a)\):\[ \frac{f(b) - f(a)}{b - a} = \frac{2.5 - \frac{5}{2}}{2 - \frac{1}{2}} = \frac{2.5 - 2.5}{1.5} = 0 \]
04

Compute the Derivative \(f^{ ext{'} ext{'}}(x)\)

Compute the derivative of \(f(x) = x + \frac{1}{x}\):\[ f'(x) = 1 - \frac{1}{x^2} \]
05

Set \(f^{'}(c) = 0\) and Solve for \(c\)

Set the derivative equal to the slope found earlier: \[ 1 - \frac{1}{c^2} = 0 \]Solve for \(c\): Rearrange: \( \frac{1}{c^2} = 1 \)\( c^2 = 1 \)\( c = \pm 1 \)Since \(c\) must lie in \(\left( \frac{1}{2}, 2 \right)\), choose \(c = 1\).
06

Verify \(c\) Lies in the Interval

Check if \(c = 1\) falls within the interval \( \left( \frac{1}{2}, 2 \right) \). Since \(c = 1\) is within the interval, it satisfies the conditions of 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!

Key Concepts

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

Differentiability
Differentiability is a fundamental concept in calculus. It tells us whether a function is smooth or has any breaks. For a function to be differentiable at a point, its derivative must exist at that point. This means the function has a defined slope, and there are no sharp corners or breaks.

There are some key aspects of differentiability:
  • A function must be continuous at a point to be differentiable there. But continuity alone does not guarantee differentiability.
  • If a function is differentiable over an interval, it means you can find a tangent line at any point in that interval.
  • In our problem, the function \(f(x) = x + \frac{1}{x}\) is differentiable over the interval \((\frac{1}{2}, 2)\).
Understanding differentiability ensures that concepts like the derivative and Mean Value Theorem are applicable. It helps in figuring out where a function's rate of change is well-defined.
Secant Line
A secant line is simply a line that passes through two points on a curve. It gives a measure of the average rate of change between these points. This concept is crucial when we deal with the Mean Value Theorem.

Here鈥檚 how the secant line works:
  • Its slope is computed as the difference in function values divided by the difference in x-values: \( \frac{f(b) - f(a)}{b - a} \).
  • The slope represents the average rate of change over the interval \([a, b]\).
In the problem, our function \(f(x) = x + \frac{1}{x}\) has a secant line over the interval \([\frac{1}{2}, 2]\) with a calculated slope of 0. This means that somewhere between \(x = \frac{1}{2}\) and \(x = 2\), the function has a point where its instantaneous rate of change equals this average rate, thanks to the Mean Value Theorem.
Derivatives
Derivatives are one of the most vital concepts in calculus, representing how a function changes at any point. The derivative of a function gives the slope of the tangent line at any specific point.

Here鈥檚 what you need to know about derivatives:
  • The derivative \(f'(x)\) of a function \(f(x)\) lets us understand its behavior 鈥� for instance, its increasing or decreasing nature.
  • Computing the derivative involves differentiating the function. For \(f(x) = x + \frac{1}{x}\), the derivative is \(f'(x) = 1 - \frac{1}{x^2}\).
  • By setting \(f'(c)\) equal to the slope of the secant line, we find the particular value \(c\) where the Mean Value Theorem holds true.
In our specific problem, solving \(1 - \frac{1}{c^2} = 0\) gives us \(c = 1\). This value falls in the interval \((\frac{1}{2}, 2)\), confirming that the derivative equaling the secant slope condition holds within the allowable range.

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

Let \(f\) be a function defined on an interval \([a, b] .\) What conditions could you place on \(f\) to guarantee that $$\min f^{\prime} \leq \frac{f(b)-f(a)}{b-a} \leq \max f^{\prime}$$ where min \(f^{\prime}\) and max \(f^{\prime}\) refer to the minimum and maximum values of \(f^{\prime}\) on \([a, b] ?\) Give reasons for your answers.

The derivative \(d t / d x\) in Example 4 a. Show that $$f(x)=\frac{x}{\sqrt{a^{2}+x^{2}}}$$ is an increasing function of \(x .\) b. Show that $$g(x)=\frac{d-x}{\sqrt{b^{2}+(d-x)^{2}}}$$ is a decreasing function of \(x .\) c. Show that $$\frac{d t}{d x}=\frac{x}{c_{1} \sqrt{a^{2}+x^{2}}}-\frac{d-x}{c_{2} \sqrt{b^{2}+(d-x)^{2}}}$$ is an increasing function of \(x .\)

Identify the inflection points and local maxima and minima of the functions graphed. Identify the intervals on which the functions are concave up and concave down. $$y=\frac{3}{4}\left(x^{2}-1\right)^{2 / 3}$$

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=x^{4}+2 x^{3}=x^{3}(x+2)$$

Find the inflection points (if any) on the graph of the function and the coordinates of the points on the graph where the function has a local maximum or local minimum value. Then graph the function in a region large enough to show all these points simultaneously. Add to your picture the graphs of the function's first and second derivatives. How are the values at which these graphs intersect the \(x\)-axis related to the graph of the function? In what other ways are the graphs of the derivatives related to the graph of the function? \(y=\frac{x^{4}}{4}-\frac{x^{3}}{3}-4 x^{2}+12 x+20\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Probability and 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.