/*! 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 13 Use the first derivative to find... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 13

Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither. $$f(x)=x^{3}-3 x+10$$

Short Answer

Expert verified
Critical points: \(x = -1\) (max), \(x = 1\) (min). Inflection point: \(x = 0\).

Step by step solution

01

Find the first derivative

To begin finding critical points, calculate the first derivative of the function \(f(x)=x^3-3x+10\). The first derivative is given by: \[ f'(x) = \frac{d}{dx}(x^3 - 3x + 10) = 3x^2 - 3. \]
02

Set the first derivative to zero

Critical points occur where the derivative is zero or undefined. Set the first derivative \(3x^2 - 3\) equal to zero and solve for \(x\): \[ 3x^2 - 3 = 0. \] Simplifying, we have: \[ x^2 = 1. \] Thus, \(x = \pm 1\).
03

Step 3:

The critical points are \(x = 1\) and \(x = -1\).
04

Use the second derivative to classify critical points

Calculate the second derivative of the function \(f(x)\): \[ f''(x) = \frac{d}{dx}(3x^2 - 3) = 6x. \] Substitute \(x = 1\) and \(x = -1\) into the second derivative to determine concavity: - For \(x = 1\), \(f''(1) = 6(1) = 6\), which is positive, indicating a local minimum. - For \(x = -1\), \(f''(-1) = 6(-1) = -6\), which is negative, indicating a local maximum.
05

Find the second derivative to determine inflection points

An inflection point occurs where the second derivative changes sign. Set \(f''(x) = 6x = 0\). Solving for \(x\) gives \(x = 0\). Substitute values around \(x = 0\) to confirm the change in sign: - For \(x < 0\), \(f''(x) < 0\). - For \(x > 0\), \(f''(x) > 0\). This sign change confirms an inflection point at \(x = 0\).
06

Graph the function to confirm critical points

Graph the function \(f(x) = x^3 - 3x + 10\). The graph should confirm: - A local minimum at \(x = 1\) (\(f(x)\) changes from decreasing to increasing). - A local maximum at \(x = -1\) (\(f(x)\) changes from increasing to decreasing). - An inflection point at \(x = 0\) (concavity changes from downward to upward).

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
The first derivative of a function gives us valuable information about the slope of the tangent line at any given point on the function's graph. For the function \( f(x) = x^3 - 3x + 10 \), the first derivative is calculated to be \( f'(x) = 3x^2 - 3 \). This derivative tells us how the function is increasing or decreasing at each point.
  • The critical points are found by setting the first derivative \( f'(x) \) equal to zero. This gives us the equations where the slope of the tangent is horizontal, indicating possible peaks, troughs, or points of inflection.
  • For this particular function, solving \( 3x^2 - 3 = 0 \) yields \( x = 1 \) and \( x = -1 \) as critical points.
Determining these points is the first step in analyzing the behavior of the function.
Second Derivative
The second derivative provides insight into the concavity of a function, indicating whether it curves upwards or downwards at a given point. For \( f(x) = x^3 - 3x + 10 \), the second derivative \( f''(x) = 6x \) helps us classify critical points further.
  • A positive second derivative, such as \( f''(1) = 6 \), suggests the function is concave up at \( x = 1 \), indicating a local minimum.
  • A negative second derivative, like \( f''(-1) = -6 \), indicates the function is concave down at \( x = -1 \), suggesting a local maximum.
This information helps us determine how the function behaves around critical points.
Inflection Points
Inflection points occur where a function changes concavity, from concave up to concave down or vice versa. These points are found by examining where the second derivative changes sign. For \( f(x) = x^3 - 3x + 10 \), setting the second derivative \( f''(x) = 6x \) equal to zero identifies \( x = 0 \) as a potential inflection point.
To confirm:
  • Check values just below and above \( x = 0 \) to detect a sign change in \( f''(x) \).
  • If \( f''(x) \) changes from negative to positive, the concavity transitions from downwards to upwards, confirming an inflection point at \( x = 0 \).
Identifying inflection points adds a deeper understanding of the function's graph.
Local Maximum
A local maximum is where a function reaches a peak value in a certain region, higher than any other nearby values. At \( x = -1 \), the function \( f(x) = x^3 - 3x + 10 \) has a local maximum.
  • This is determined both by setting \( f'(x) = 0 \) and confirming with the second derivative test.
  • Since \( f''(-1) = -6 \) is less than zero, it indicates concave down, making \( x = -1 \) a local maximum.
      • The function decreases before and increases after this point, confirming it's a peak in the local region.
Local Minimum
A local minimum occurs at a point where the function has a dip, lower than any nearby values. For the function \( f(x) = x^3 - 3x + 10 \), \( x = 1 \) is a local minimum.
  • First, we set \( f'(x) = 0 \) to identify possible minima.
  • The second derivative, \( f''(1) = 6 \), is positive, indicating concave up at this point.
This suggests \( x = 1 \) is where the function reaches a trough, with values rising before and after, marking it clearly as a local minimum.

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

If you have 100 feet of fencing and want to enclose a rectangular area up against a long, straight wall, what is the largest area you can enclose?

A curve representing the total number of people, \(P\), infected with a virus often has the shape of a logistic curve of the form $$P=\frac{L}{1+C e^{-k t}}$$ with time \(t\) in weeks. Suppose that 10 people originally have the virus and that in the early stages the number of people infected is increasing approximately exponentially, with a continuous growth rate of \(1.78 .\) It is estimated that, in the long run, approximately 5000 people will become infected. (a) What should we use for the parameters \(k\) and \(L ?\) (b) Use the fact that when \(t=0,\) we have \(P=10,\) to find \(C\) (c) Now that you have estimated \(L, k,\) and \(C,\) what is the logistic function you are using to model the data? Graph this function. (d) Estimate the length of time until the rate at which people are becoming infected starts to decrease. What is the value of \(P\) at this point?

In a chemical reaction, substance \(A\) combines with substance \(B\) to form substance \(Y .\) At the start of the reaction, the quantity of \(A\) present is \(a\) grams, and the quantity of \(B\) present is \(b\) grams. At time \(t\) seconds after the start of the reaction, the quantity of \(Y\) present is \(y\) grams. Assume \(a< b\) and \(y \leq a .\) For certain types of reactions, the rate of the reaction, in grams/sec, is given by Rate \(=k(a-y)(b-y), \quad k\) is a positive constant. (a) For what values of \(y\) is the rate nonnegative? Graph the rate against \(y\) (b) Use your graph to find the value of \(y\) at which the rate of the reaction is fastest.

A demand function is \(p=400-2 q,\) where \(q\) is the quantity of the good sold for price \(\$ p\) (a) Find an expression for the total revenue, \(R\), in terms of \(q\) (b) Differentiate \(R\) with respect to \(q\) to find the marginal revenue, \(M R,\) in terms of \(q .\) Calculate the marginal revenue when \(q=10\) (c) Calculate the change in total revenue when production increases from \(q=10\) to \(q=11\) units. Confirm that a one-unit increase in \(q\) gives a reasonable approximation to the exact value of \(M R\) obtained in part (b).

A company estimates that the total revenue, \(R\), in dollars, received from the sale of \(q\) items is \(R=\ln \left(1+1000 q^{2}\right)\) Calculate and interpret the marginal revenue if \(q=10\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

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.