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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 44
Chapter 4, Problem 44

Each of Exercises 43–48 gives the first derivative of a function y = ƒ(𝓍). (a) At what points, if any, does the graph of ƒ have a local maximum, local minimum, or inflection point? (b) Sketch the general shape of the graph.
y' = 𝓍² ― 𝓍―6

Verified step by step guidance
1
First, identify the critical points by setting the first derivative y' = 𝓍² - 𝓍 - 6 equal to zero and solving for 𝓍. This will help us find potential points for local maxima, minima, or inflection points.
Factor the quadratic equation 𝓍² - 𝓍 - 6 = 0 to find the values of 𝓍. The equation can be factored as (𝓍 - 3)(𝓍 + 2) = 0, giving the critical points 𝓍 = 3 and 𝓍 = -2.
To determine whether these critical points are local maxima or minima, use the second derivative test. Compute the second derivative y'' by differentiating y' = 𝓍² - 𝓍 - 6 again, resulting in y'' = 2𝓍 - 1.
Evaluate the second derivative at the critical points: y''(3) and y''(-2). If y''(𝓍) > 0, the point is a local minimum; if y''(𝓍) < 0, the point is a local maximum. If y''(𝓍) = 0, further analysis is needed to determine the nature of the point.
To find inflection points, set the second derivative y'' = 2𝓍 - 1 equal to zero and solve for 𝓍. This will indicate where the concavity of the function changes, which is a characteristic of inflection points.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

First Derivative Test

The First Derivative Test is a method used to determine local extrema of a function. By analyzing the sign of the first derivative, we can identify intervals where the function is increasing or decreasing. A local maximum occurs where the derivative changes from positive to negative, while a local minimum occurs where it changes from negative to positive.
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The First Derivative Test: Finding Local Extrema

Critical Points

Critical points are values of x where the first derivative is either zero or undefined. These points are essential for finding local maxima, minima, and inflection points. In the given derivative, we find critical points by solving the equation y' = 0, which helps us identify where the function's behavior changes.
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Critical Points

Inflection Points

Inflection points occur where the concavity of a function changes, which can be determined by analyzing the second derivative. However, in the context of the first derivative, inflection points can also be inferred from changes in the behavior of the first derivative itself. Identifying these points helps in sketching the overall shape of the graph.
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Related Practice
Textbook Question

Each of Exercises 43–48 gives the first derivative of a function y = ƒ(𝓍). (a) At what points, if any, does the graph of ƒ have a local maximum, local minimum, or inflection point? (b) Sketch the general shape of the graph.

y' = 𝓍⁴ ― 2𝓍² 

Textbook Question

Finding Position from Velocity or Acceleration


Exercises 41–44 give the velocity v = ds/dt and initial position of an object moving along a coordinate line. Find the object’s position at time t.


v = sin πt, s(0) = 0

Textbook Question

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.

46. y = cos(x) + √3 * sin(x), 0 ≤ x ≤ 2π

Textbook Question

Finding Position from Velocity or Acceleration


Exercises 45–48 give the acceleration a=d²s/dt², initial velocity, and initial position of an object moving on a coordinate line. Find the object’s position at time t.


a = 32, v(0) = 20, s(0) = 5

Textbook Question

Graphs and Graphing


Graph the curves in Exercises 33–42.

______

y = 𝓍√4 ― 𝓍²

Textbook Question

Finding Position from Velocity or Acceleration


Exercises 41–44 give the velocity v = ds/dt and initial position of an object moving along a coordinate line. Find the object’s position at time t.


v = 9.8t + 5, s(0) = 10