Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

5. Graphical Applications of Derivatives

The First Derivative Test

Calculus

5. Graphical Applications of Derivatives

The First Derivative Test

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1:37 minutes

Problem 4.3.3a

Textbook Question

Analyzing Functions from Derivatives

Answer the following questions about the functions whose derivatives are given in Exercises 1–14:

a. What are the critical points of f?

f′(x) = (x − 1)²(x + 2)

Verified step by step guidance

To find the critical points of the function f, we need to determine where its derivative f'(x) is equal to zero or undefined. In this case, f'(x) = (x - 1)²(x + 2).

Set the derivative equal to zero: (x - 1)²(x + 2) = 0. This equation will help us find the values of x where the derivative is zero, indicating potential critical points.

Solve the equation (x - 1)²(x + 2) = 0 by setting each factor equal to zero: (x - 1)² = 0 and (x + 2) = 0.

For (x - 1)² = 0, solve for x to get x = 1. This is a critical point because the derivative is zero at this value.

For (x + 2) = 0, solve for x to get x = -2. This is another critical point because the derivative is zero at this value.

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Key Concepts

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

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points are significant because they can indicate local maxima, minima, or points of inflection. To find critical points, set the derivative equal to zero and solve for the variable, or identify where the derivative does not exist.

Factoring Polynomials

Factoring polynomials involves expressing a polynomial as a product of its factors. This is crucial for solving equations where the polynomial is set to zero, as it allows us to find the roots or solutions. In the given derivative, f′(x) = (x − 1)²(x + 2), factoring helps identify the values of x that make the derivative zero.

Derivative Analysis

Analyzing the derivative of a function provides insights into the function's behavior, such as increasing or decreasing intervals and concavity. The sign and value of the derivative indicate the slope of the tangent line at any point on the function. For f′(x) = (x − 1)²(x + 2), examining the derivative helps determine where the function changes direction or has potential extrema.

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In Exercises 53 and 54, show that the function has neither an absolute minimum nor an absolute maximum on its natural domain.

y = x¹¹ + x³ + x − 5

Textbook Question

Theory and Examples

Maximum height of a vertically moving body The height of a body moving vertically is given by s = −12gt² + υ₀t + s₀, g > 0, with s in meters and t in seconds. Find the body’s maximum height.

Textbook Question

Analyzing Functions from Derivatives