/*! 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 20 In Exercises \(13-20\) , find al... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 20

In Exercises \(13-20\) , find all points of inflection of the function. $$y=\frac{x}{x^{2}+1}$$

Short Answer

Expert verified
The inflection points of the given function are at \((0,0)\), \((-1, -0.5)\), and \((1, 0.5)\)

Step by step solution

01

Compute the first derivative

Using the quotient rule, the first derivative of \(y\) with respect to \(x\) is given by \(y'=\frac{(x^{2}+1)\cdot1-x\cdot2x}{(x^{2}+1)^2}=\frac{1-x^{2}}{(x^{2}+1)^2}\)
02

Find the second derivative

Again applying the quotient rule,\[y''=\frac{(x^{2}+1)^2\cdot-2x-(1-x^{2})\cdot2(x^{2}+1)\cdot2x}{(x^{2}+1)^4}=\frac{-4x^{3}-4x}{(x^{2}+1)^4}\]
03

Find the inflection points

To find the points of inflection, set \(y''=0\), we get \(\frac{-4x^{3}-4x}{(x^{2}+1)^4}=0\)Solve for \(x\), we then solve the numerator to get the solutions as \(x = 0\) and \(x = -1, 1\) for the equation \(-4x^{3}-4x = 0\). So, we have three critical points. We then substitute these values into the original expression to find the corresponding 'y' values. The inflection points hence are \((0,0)\), \((-1, -0.5)\), and \((1, 0.5)\).

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.

Quotient Rule
The quotient rule is a technique for finding the derivative of a function that is the ratio of two differentiable functions. If you have a function in the form \(\frac{u(x)}{v(x)}\), you can use the quotient rule to take its derivative:
  • Let \(y = \frac{u(x)}{v(x)}\).
  • The first derivative \(y'\) is given by \(y' = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}\).
This rule simplifies the process of handling complex expressions, particularly when both the numerator and denominator are polynomials or other functions that can be differentiated separately. In our exercise, \(u(x) = x\) and \(v(x) = x^2 + 1\), and their derivatives are very straightforward to calculate.

This foundational rule is crucial for ensuring accuracy when deriving functions involving both multiplication and division.
First Derivative
The first derivative of a function gives us the rate at which the function is changing at any point and helps identify critical behaviors like turning points or steep slopes. In this exercise, we found the first derivative of \(y = \frac{x}{x^2+1}\) using the quotient rule.

We computed:
  • The derivative of the numerator \(u'(x) = 1\) and the denominator \(v(x) = 2x\).
  • This resulted in the expression for \(y'\), the first derivative, as \(y' = \frac{1-x^2}{(x^2+1)^2}\).
The first derivative functions as a powerful tool to examine where the slope of the tangent line is zero, which indicates potential maximum or minimum points on the graph of the original function.
Second Derivative
Finding the second derivative is essential for understanding the concavity of a function, which tells us how the slope of the tangent line is changing. A key outcome of this is the determination of inflection points, where the concavity changes direction.

For our function, we arrived at the second derivative \(y''\) through a second application of the quotient rule. After finding it as \(y'' = \frac{-4x^3-4x}{(x^2+1)^4}\), it poses as a critical step towards examining concavity:
  • Positive \(y''\) at a point suggests the curve is concave up, indicating a local minimum.
  • Negative \(y''\) indicates concave down, suggesting a local maximum.
  • When \(y''\) transitions through zero, there might be an inflection point.
Knowing when to employ the second derivative is instrumental in a more detailed analysis of the function鈥檚 graph.
Critical Points
Critical points occur where the derivative of a function is zero or undefined, shedding light on possible local maxima, minima, or points of inflection. These are particularly key in turning the data derived from derivatives into insights about the behavior of our original function.

To find inflection points, we set the second derivative equal to zero and solved for \(x\), resulting in the critical points \(x = 0\) and \(x = \pm1\).
  • These points indicate where the function \(y = \frac{x}{x^2+1}\) possibly changes concavity, moving from concave up to concave down or vice versa.
  • After calculating the corresponding \(y\) values, we identified the points of inflection as \((0,0), (-1,-0.5), (1,0.5)\).
This analysis elucidates significant moments in the function, necessary for both purely mathematical investigations and practical real-world interpretations.

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

You may use a graphing calculator to solve the following problems. True or False If the radius of a circle is expanding at a constant rate, then its circumference is increasing at a constant rate. Justify your answer.

Draining Hemispherical Reservoir Water is flowing at the rate of 6 \(\mathrm{m}^{3} / \mathrm{min}\) from a reservoir shaped like a hemispherical bowl of radius \(13 \mathrm{m},\) shown here in profile. Answer the following questions given that the volume of water in a hemispherical bowl of radius \(R\) is \(V=(\pi / 3) y^{2}(3 R-y)\) when the water is \(y\) units deep. (a) At what rate is the water level changing when the water is 8 m deep? (b) What is the radius \(r\) of the water's surface when the water is \(y\) m deep? (c) At what rate is the radius \(r\) changing when the water is 8 \(\mathrm{m}\) deep?

Draining Conical Reservoir Water is flowing at the rate of 50 \(\mathrm{m}^{3} / \mathrm{min}\) from a concrete conical reservoir (vertex down) of base radius 45 \(\mathrm{m}\) and height 6 \(\mathrm{m} .\) (a) How fast is the water level falling when the water is 5 \(\mathrm{m}\) deep? (b) How fast is the radius of the water's surface changing at that moment? Give your answer in \(\mathrm{cm} / \mathrm{min.}\)

Estimating Volume You can estimate the volume of a sphere by measuring its circumference with a tape measure, dividing by 2\(\pi\) to get the radius, then using the radius in the volume formula. Find how sensitive your volume estimate is to a 1\(/ 8\) in. error in the circumference measurement by filling in the table below for spheres of the given sizes. Use differentials when filling in the last column. $$\begin{array}{|c|c|c|}\hline \text { Sphere Type } & {\text { True Radius }} & {\text { Tape Error Radius Error Volume Error }} \\ \hline \text { Orange } & {2 \text { in. }} & {1 / 8 \text { in. }} \\ \hline \text { Melon } & {4 \text { in. }} & {1 / 8 \text { in. }} \\ \hline \text { Beach Ball } & {7 \text { in. }} & {1 / 8 \text { in. }}\end{array}$$

Free Fall On the moon, the acceleration due to gravity is 1.6 \(\mathrm{m} / \mathrm{sec}^{2} .\) (a) If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 30 sec later? (b) How far below the point of release is the bottom of the crevasse? (c) If instead of being released from rest, the rock is thrown into the crevasse from the same point with a downward velocity of \(4 \mathrm{m} / \mathrm{sec},\) when will it hit the bottom and how fast will it be going when it does?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

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.