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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 4.3.5a
Chapter 4, Problem 4.3.5a

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)(x + 2)(x − 3)

Verified step by step guidance
1
To find the critical points of the function f, we need to determine where the derivative f'(x) is equal to zero or undefined. Since f'(x) = (x - 1)(x + 2)(x - 3), we focus on where this expression equals zero.
Set f'(x) = 0: (x - 1)(x + 2)(x - 3) = 0. This equation is satisfied when any of the factors is zero.
Solve each factor for x: x - 1 = 0, x + 2 = 0, and x - 3 = 0. This gives us the potential critical points.
The solutions to these equations are x = 1, x = -2, and x = 3. These are the critical points of the function f.
Verify that these points are indeed critical points by ensuring that f'(x) changes sign around these values, indicating a change in the behavior of the function f.

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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, which in this case involves solving f′(x) = (x − 1)(x + 2)(x − 3) = 0.
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Critical Points

Factoring Polynomials

Factoring polynomials is a method used to simplify expressions and solve equations. It involves expressing a polynomial as a product of its factors. For the derivative f′(x) = (x − 1)(x + 2)(x − 3), the factors are already given, making it straightforward to find the roots by setting each factor equal to zero: x = 1, x = -2, and x = 3.
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Introduction to Polynomial Functions

Roots of Equations

The roots of an equation are the values of the variable that satisfy the equation, making it equal to zero. In the context of derivatives, finding the roots of f′(x) helps identify the critical points of the function f(x). For f′(x) = (x − 1)(x + 2)(x − 3), the roots are x = 1, x = -2, and x = 3, which are the critical points of the function.
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Related Practice
Textbook Question

Finding Antiderivatives

In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.

(2/3)x⁻¹ᐟ³

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Textbook Question

51. Frictionless cart A small frictionless cart, attached to the wall by a spring, is pulled 10 cm from its rest position and released at time t = 0 to roll back and forth for 4 sec. Its position at time t is s = 10 cos πt.

a. What is the cart's maximum speed? When is the cart moving that fast? Where is it then? What is the magnitude of the acceleration then?

Textbook Question

Applications


Suppose that f(x) = d/dx (1 − √x) and g(x) = d/dx (x + 2).


Find:


∫f(x) dx

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)

Textbook Question

Finding Antiderivatives

In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.

2x

1
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Textbook Question

Identifying Extrema


In Exercises 41–52:


a. Identify the function’s local extreme values in the given domain, and say where they occur.


f(t) = t³ − 3t², −∞ < t ≤ 3