/*! 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 74 Find the relative maxima and rel... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 74

Find the relative maxima and relative minima, if any, of each function. $$ f(x)=x^{2} e^{-x} $$

Short Answer

Expert verified
The function \(f(x) = x^2 e^{-x}\) has a relative maximum at \(x = 0\) and a relative minimum at \(x = 2\).

Step by step solution

01

Find the first derivative of the function

To find the first derivative of the function \(f(x)\), we'll need to use the product rule. The product rule states that if we have a function \(f(x) = g(x)h(x)\), then the derivative of \(f(x)\) is: $$ f'(x) = g'(x)h(x) + g(x)h'(x) $$ Let \(g(x) = x^2\) and \(h(x) = e^{-x}\). Then, the first derivative of \(f(x)\) is: $$ f'(x) = (2x)e^{-x} + x^2 (-e^{-x}) $$
02

Solve for critical points

To find the critical points of the function, we set the first derivative equal to 0 and solve for x: $$ 0 = (2x)e^{-x} + x^2 (-e^{-x}) $$ First, factor out an \(x e^{-x}\) term: $$ 0 = x e^{-x} (2 - x) $$ This yields two critical points: \(x_1 = 0\) and \(x_2 = 2\).
03

Find the second derivative of the function

To determine the nature of the critical points, we'll use the second derivative test. We need to find the second derivative of \(f(x)\). To do this, we'll find the derivative of \(f'(x)\): $$ f''(x) = (-2)e^{-x} + 2xe^{-x} + 2x(-e^{-x}) - x^2 e^{-x} $$
04

Evaluate the second derivative at the critical points

Next, we substitute the critical points into the second derivative: At \(x = 0\): $$ f''(0) = (-2)e^0 + 0 = -2 < 0 $$ At \(x = 2\): $$ f''(2) = (-2)e^{-2} + 4e^{-2} - 4e^{-2} + 4e^{-2} = 2e^{-2} > 0 $$
05

Determine the nature of critical points using the second derivative test

Now, let's interpret our results: - \(f''(0) < 0\), which means that \(x = 0\) is a critical point with a relative maximum. - \(f''(2) > 0\), which means that \(x = 2\) is a critical point with a relative minimum. Therefore, the function \(f(x) = x^2 e^{-x}\) has a relative maximum at \(x = 0\) and a relative minimum at \(x = 2\).

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.

Critical Points
Critical points of a function are specific values of the variable, usually denoted as \( x \), where the function's derivative is either zero or undefined. These points hold great importance because they are potential locations where the function changes direction, which can include peaks, valleys, or even plateaus.

Here's a quick way to identify them:
  • Set the first derivative of the function, \( f'(x) \), equal to zero.
  • Factor the equation if possible to solve for all possible values of \( x \) where the derivative equals zero.
For example, in our exercise with \( f(x) = x^2 e^{-x} \), we found the critical points by setting \( f'(x) = 0 \) and solving for \( x \). This led us to the critical points \( x_1 = 0 \) and \( x_2 = 2 \). These are places where the slope of the curve changes, indicating a local maximum or minimum.
First Derivative
The first derivative of a function, denoted as \( f'(x) \), is a tool used to understand how the function is changing at any given point. It represents the rate of change, similar to velocity, but for functions. In simpler terms, the first derivative tells us how steep the graph of the function is at any point, and in what direction it鈥檚 sloping.

To compute the first derivative:
  • Apply differentiation rules, like the product rule, chain rule, or quotient rule, depending on the function's form.
  • For the given function, \( f(x) = x^2 e^{-x} \), we used the product rule to differentiate, resulting in \( f'(x) = (2x)e^{-x} + x^2 (-e^{-x}) \).
The first derivative is crucial because setting it to zero helps locate the critical points where potential maxima or minima occur.
Second Derivative Test
After identifying the critical points using the first derivative, the second derivative test helps determine the type of extrema each critical point signifies.

Here's how the second derivative test is conducted:
  • Find the second derivative of the function, denoted as \( f''(x) \).
  • Substitute each critical point into \( f''(x) \).
  • Analyze the value of the second derivative:
    • If \( f''(x) > 0 \), the function has a relative minimum at that point.
    • If \( f''(x) < 0 \), the function has a relative maximum.
    • If \( f''(x) = 0 \), the test is inconclusive.
In this exercise, we evaluated the second derivative at \( x = 0 \) and \( x = 2 \):
  • At \( x = 0 \), \( f''(0) = -2 \), signaling a relative maximum.
  • At \( x = 2 \), \( f''(2) = 2e^{-2} \) > 0, indicating a relative minimum.
Thus, the second derivative test confirms the nature of the critical points found earlier.

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

Find the absolute maximum value and the absolute minimum value, if any, of each function. $$ f(x)=x^{2 / 3}\left(x^{2}-4\right) \text { on }[-1,2] $$

Neilsen Cookie Company sells its assorted butter cookies in containers that have a net content of \(1 \mathrm{lb}\). The estimated demand for the cookies is \(1,000,000\) 1-lb containers. The setup cost for each production run is $$\$ 500$$, and the manufacturing cost is $$\$ .50$$ for each container of cookies. The cost of storing each container of cookies over the year is $$\$ .40$$. Assuming uniformity of demand throughout the year and instantaneous production, how many containers of cookies should Neilsen produce per production run in order to minimize the production cost? Hint: Following the method of Example 5 , show that the total production cost is given by the function $$ C(x)=\frac{500,000,000}{x}+0.2 x+500,000 $$ Then minimize the function \(C\) on the interval \((0,1,000,000)\).

AvERAGE SPEED OF A VEHICLE The average speed of a vehicle on a stretch of Route 134 between 6 a.m. and 10 a.m. on a typical weekday is approximated by the function $$ f(t)=20 t-40 \sqrt{t}+50 \quad(0 \leq t \leq 4) $$ where \(f(t)\) is measured in miles per hour and \(t\) is measured in hours, with \(t=0\) corresponding to 6 a.m. At what time of the morning commute is the traffic moving at the slowest rate? What is the average speed of a vehicle at that time?

The total worldwide box-office receipts for a long-running movie are approximated by the function $$ T(x)=\frac{120 x^{2}}{x^{2}+4} $$ where \(T(x)\) is measured in millions of dollars and \(x\) is the number of years since the movie's release. Sketch the graph of the function \(T\) and interpret your results.

What are the dimensions of a closed rectangular box that has a square cross section, a capacity of 128 in. \(^{3}\), and is constructed using the least amount of material?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Discrete 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.