/*! 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 26 Find all relative extrema. Use t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 26

Find all relative extrema. Use the Second Derivative Test where applicable. \(f(x)=\sqrt{x^{2}+1}\)

Short Answer

Expert verified
The function \(f(x)=\sqrt{x^{2}+1}\) has one relative minimum at \(x=0\).

Step by step solution

01

Differentiation

Firstly, we need to calculate the derivative of the function. A useful formula for this task is the chain rule. For a function given \(h(u)=\sqrt{u}\), its derivative is \(\frac{1}{2\sqrt{u}}\cdot u'\), so we have \(f'(x)=\frac{x}{\sqrt{x^{2}+1}}\).
02

Find the Critical Points

The critical points are the values of the variable where the function has a local maximum or minimum. These critical points can be found by setting the derivative equal to zero and solve for \(x\), which gives \(\frac{x}{\sqrt{x^{2}+1}}=0\), and hence, \(x = 0\) is the only critical point of the function.
03

Second Derivative

Next, we calculate the second derivative. Calculating the second derivative, \(f''(x)\), is a bit complex and requires use of the quotient rule and chain rule again. After some simplification, we get \(f''(x)=\frac{1}{\left(x^{2}+1\right)^{\frac{3}{2}}}\).
04

Apply the Second Derivative Test

The Second Derivative Test involves substituting the critical points into the second derivative. If the value we get is positive, the function has a relative minimum at that point, if it's negative, a relative maximum. For the point \(x=0\), we have \(f''(0) = 1\), which is greater than zero.
05

Identify and Label Extrema

Since the second derivative at \(x=0\) is positive, the function \(f(x)\) has a relative minimum at \(x=0\) by the Second Derivative Test.

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

In Exercises 87 and \(88,\) (a) use a graphing utility to graph \(f\) and \(g\) in the same viewing window, (b) verify algebraically that \(f\) and \(g\) represent the same function, and (c) zoom out sufficiently far so that the graph appears as a line. What equation does this line appear to have? (Note that the points at which the function is not continuous are not readily seen when you zoom out.) $$ \begin{array}{l} f(x)=-\frac{x^{3}-2 x^{2}+2}{2 x^{2}} \\ g(x)=-\frac{1}{2} x+1-\frac{1}{x^{2}} \end{array} $$

Use the definitions of increasing and decreasing functions to prove that \(f(x)=x^{3}\) is increasing on \((-\infty, \infty)\).

Let \(f\) and \(g\) represent differentiable functions such that \(f^{\prime \prime} \neq 0\) and \(g^{\prime \prime} \neq 0\). Prove that if \(f\) and \(g\) are positive, increasing, and concave upward on the interval \((a, b),\) then \(f g\) is also concave upward on \((a, b)\).

A ball bearing is placed on an inclined plane and begins to roll. The angle of elevation of the plane is \(\theta .\) The distance (in meters) the ball bearing rolls in \(t\) seconds is \(s(t)=4.9(\sin \theta) t^{2}\) (a) Determine the speed of the ball bearing after \(t\) seconds. (b) Complete the table and use it to determine the value of \(\theta\) that produces the maximum speed at a particular time. $$ \begin{array}{|l|l|l|l|l|l|l|l|} \hline \boldsymbol{\theta} & 0 & \pi / 4 & \pi / 3 & \pi / 2 & 2 \pi / 3 & 3 \pi / 4 & \pi \\ \hline \boldsymbol{s}^{\prime}(\boldsymbol{t}) & & & & & & & \\ \hline \end{array} $$

The range \(R\) of a projectile fired with an initial velocity \(v_{0}\) at an angle \(\theta\) with the horizontal is \(R=\frac{v_{0}^{2} \sin 2 \theta}{g},\) where \(g\) is the acceleration due to gravity. Find the angle \(\theta\) such that the range is a maximum.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Calculus

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.