Textbook Question
Second Derivative Test Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima.
f(x) = eˣ(x - 2)²
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Ch. 4 - Applications of the Derivative
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 4, Problem 4.1.27
Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant.
ƒ(x) = 3x³ + 3x² / 2 - 2x
Verified step by step guidance1
To find the critical points of the function \( f(x) = 3x^3 + \frac{3}{2}x^2 - 2x \), we first need to find its derivative, \( f'(x) \).
Differentiate the function: \( f'(x) = \frac{d}{dx}(3x^3) + \frac{d}{dx}(\frac{3}{2}x^2) - \frac{d}{dx}(2x) \).
Calculate each derivative: \( \frac{d}{dx}(3x^3) = 9x^2 \), \( \frac{d}{dx}(\frac{3}{2}x^2) = 3x \), and \( \frac{d}{dx}(2x) = 2 \).
Combine the derivatives to get \( f'(x) = 9x^2 + 3x - 2 \).
Set the derivative equal to zero to find the critical points: \( 9x^2 + 3x - 2 = 0 \). Solve this quadratic equation for \( x \) to find the critical points.
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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 either zero or undefined. These points are essential for identifying local maxima, minima, and points of inflection. To find critical points, one typically takes the derivative of the function and solves for the values of x that satisfy the condition.
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Critical Points
Derivative
The derivative of a function measures the rate at which the function's value changes as its input changes. It is a fundamental concept in calculus, representing the slope of the tangent line to the curve at any given point. For polynomial functions, the derivative can be calculated using power rules, which simplify the process of finding critical points.
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05:44Derivatives
Polynomial Functions
Polynomial functions are expressions that involve variables raised to whole number powers, combined using addition, subtraction, and multiplication. They are characterized by their degree, which is the highest power of the variable. Understanding the behavior of polynomial functions, including their critical points and end behavior, is crucial for analyzing their graphs and determining local extrema.
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6:04Introduction to Polynomial Functions
Related Practice
Textbook Question
Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.
ƒ(x) = x/(x²+9)⁵ on [-2,2]
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Textbook Question
{Use of Tech} Finding roots with Newton’s method For the given function f and initial approximation x₀, use Newton’s method to approximate a root of f. Stop calculating approximations when two successive approximations agree to five digits to the right of the decimal point after rounding. Show your work by making a table similar to that in Example 1.
f(x) = x² - 10; x₀ = 3
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Textbook Question
{Use of Tech} Finding roots with Newton’s method For the given function f and initial approximation x₀, use Newton’s method to approximate a root of f. Stop calculating approximations when two successive approximations agree to five digits to the right of the decimal point after rounding. Show your work by making a table similar to that in Example 1.
f(x) = x ln (x + 1) -1 ; x₀ = 1.7
Textbook Question
Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.
ƒ(x) = (4x³/3) + 5x² - 6x on [0,5]
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Textbook Question
Increasing and decreasing functions. Find the intervals on which f is increasing and the intervals on which it is decreasing.
f(x) = eˣ/(e²ᵉ + 1)
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