/*! 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 38 Find the \(x\) -coordinates of a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 38

Find the \(x\) -coordinates of all critical points of the given function. Determine whether each critical point is a relative maximum, minimum, or neither by first applying the second derivative test, and, if the test fails, by some other method. \(f(t)=-2 t^{3}+3 t\)

Short Answer

Expert verified
The critical points of the function \(f(t) = -2t^3 + 3t\) are \(t = \frac{1}{\sqrt{2}}\) and \(t = -\frac{1}{\sqrt{2}}\). By applying the second derivative test, we find that the relative maximum occurs at the \(x\)-coordinate \(t = \frac{1}{\sqrt{2}}\), and the relative minimum occurs at the \(x\)-coordinate \(t = -\frac{1}{\sqrt{2}}\).

Step by step solution

01

Find the first derivative of the function

To find the critical points, we first need to find the first derivative of the function. The function is \(f(t)=-2t^3+3t\). Using the power rule, we get: \[f'(t) = \frac{d}{dt}(-2t^3 + 3t) = -6t^2 + 3\]
02

Find the critical points by setting the first derivative equal to zero

Now we need to find the critical points by setting the first derivative equal to zero: \[-6t^2 + 3 = 0\] Divide by -3 to simplify: \[2t^2 - 1 = 0\] Now solve for \(t\): \[2t^2 = 1\] \[t^2 = \frac{1}{2}\] \[t = \pm \frac{1}{\sqrt{2}}\] So the critical points are \(t = \frac{1}{\sqrt{2}}\) and \(t = -\frac{1}{\sqrt{2}}\).
03

Find the second derivative of the function

In order to apply the second derivative test, we need to find the second derivative of the function. Using the first derivative from Step 1: \[f'(t) = -6t^2 + 3\] We find the second derivative by taking the derivative of \(f'(t)\): \[f''(t) = \frac{d^2}{dt^2}(-6t^2 + 3) = -12t\]
04

Apply the second derivative test

Now we will apply the second derivative test by finding the value of the second derivative at each critical point: For \(t = \frac{1}{\sqrt{2}}\): \[f''\left(\frac{1}{\sqrt{2}}\right) = -12\left(\frac{1}{\sqrt{2}}\right) = -\frac{12}{\sqrt{2}}\] Since this value is negative, the critical point is a relative maximum. For \(t = -\frac{1}{\sqrt{2}}\): \[f''\left(-\frac{1}{\sqrt{2}}\right) = -12\left(-\frac{1}{\sqrt{2}}\right) = \frac{12}{\sqrt{2}}\] Since this value is positive, the critical point is a relative minimum.
05

Conclusion

So, the relative maximum occurs at the \(x\)-coordinate \(t = \frac{1}{\sqrt{2}}\), and the relative minimum occurs at the \(x\)-coordinate \(t = -\frac{1}{\sqrt{2}}\).

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 Test
The First Derivative Test is an essential calculus tool for determining the nature of critical points in the graph of a function. It involves analyzing the sign of the derivative before and after the critical point. This test tells us whether a critical point is a relative maximum, relative minimum, or neither. To apply this test, follow the given steps.

  • Find the first derivative of the function.
  • Identify the critical points by setting the first derivative equal to zero.
  • Check the sign of the derivative before and after each critical point. If the sign changes from positive to negative, the point is a relative maximum. If it changes from negative to positive, it's a relative minimum. If there鈥檚 no change, the test is inconclusive.
In the provided exercise, the first derivative of the function f(t) = -2t^3 + 3t is f'(t) = -6t^2 + 3, and the critical points found are t = 卤1/鈭�2. By subbing in values around these critical points into the first derivative, we can determine their nature based on the sign change.
Second Derivative Test
The Second Derivative Test is another effective method for determining the concavity of a function at its critical points and thereby identifying the relative maximums and minimums. It is particularly helpful when the First Derivative Test is inconclusive. The steps to apply the Second Derivative Test are:

  • After finding the critical points, calculate the second derivative of the function.
  • Evaluate the second derivative at each critical point.
  • If the second derivative at the critical point is positive, the function has a relative minimum there. If it鈥檚 negative, the point is a relative maximum. If it equals zero, the test fails and does not provide conclusive information about the point.
For our exercise, the second derivative f''(t) = -12t helps us classify the critical points. When evaluated at t = 1/鈭�2, we get a negative value, indicating a relative maximum. At t = -1/鈭�2, we get a positive value, indicating a relative minimum.
Relative Maximum and Minimum
The terms 'relative maximum' and 'relative minimum' describe points on the graph of a function where the function values are locally the highest or lowest, respectively. They are essential in understanding the shape and turning points of functions. The critical conditions for a relative maximum or minimum are:

  • There must be a critical point鈥攁n x-value for which the first derivative is zero or the derivative does not exist.
  • The First or Second Derivative Test must classify the point as a relative maximum or minimum.
In the context of our exercise, the critical points at t = 卤1/鈭�2 are confirmed to be relative maximum and minimum, respectively, using the Second Derivative Test. The relative maximum occurs where the graph changes from concave up to concave down, and the relative minimum occurs where the graph changes from concave down to concave up. These points are pivotal in sketching the overall behavior of the function's graph.

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

The demand for personal computers in the home goes up with household income. For a given community, we can approximate the average number of computers in a home as $$q=0.3454 \ln x-3.047 \quad 10,000 \leq x \leq 125,000$$ where \(x\) is mean household income. \({ }^{60}\) Your community has a mean income of \(\$ 30,000\), increasing at a rate of \(\$ 2,000\) per year. How many computers per household are there, and how fast is the number of computers in a home increasing? (Round your answer to four decimal places.)

A time- series study of the demand for higher education, using tuition charges as a price variable, yields the following result: $$\frac{d q}{d p} \cdot \frac{p}{q}=-0.4$$ where \(p\) is tuition and \(q\) is the quantity of higher education. \(\mathbf{2 5}\). Which of the following is suggested by the result? (A) As tuition rises, students want to buy a greater quantity \(\quad \mathbf{2 6}\). of education. (B) As a determinant of the demand for higher education, income is more important than price. (C) If colleges lowered tuition slightly, their total tuition receipts would increase. (D) If colleges raised tuition slightly, their total tuition receipts would increase. (E) Colleges cannot increase enrollments by offering larger scholarships.

The percentage of U.S.-issued mortgages that are subprime can be approximated by $$A(t)=\frac{15.0}{1+8.6(1.8)^{-t}} \text { percent } \quad(0 \leq t \leq 8)$$ \(t\) years after the start of \(2000^{37}\) Graph the function as well as its first and second derivatives. Determine, to the nearest whole number, the values of \(t\) for which the graph of \(A\) is concave up and concave down, and the \(t\) -coordinate of any points of inflection. What does the point of inflection tell you about subprime mortgages?

Daily oil imports to the United States from Mexico can be approximated by \(q(t)=-0.015 t^{2}+0.1 t+1.4\) million barrels \((0 \leq t \leq 8)\) where \(t\) is time in years since the start of \(2000 .^{56}\) At the start of 2004 the price of oil was \(\$ 30\) per barrel and increasing at a rate of \(\$ 40\) per year. \({ }^{57}\) How fast was (daily) oil expenditure for imports from Mexico changing at that time?

Rewrite the statements and questions in mathematical notation. The population \(P\) is currently 10,000 and growing at a rate of 1,000 per year.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Pure 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.