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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 4 - Applications of the DerivativeProblem 19
Chapter 4, Problem 19

Use ƒ' and ƒ" to complete parts (a) and (b). 


a. Find the intervals on which f is increasing and the intervals on which it is decreasing.


b. Find the intervals on which f is concave up and the intervals on which it is concave down.


ƒ(x) = x⁹/9 + 3x⁵ - 16x

Verified step by step guidance
1
First, find the first derivative ƒ'(x) of the function ƒ(x) = \( \frac{x^9}{9} + 3x^5 - 16x \). Use the power rule for differentiation: \( \frac{d}{dx} x^n = nx^{n-1} \).
Calculate ƒ'(x): \( \frac{d}{dx} \left( \frac{x^9}{9} \right) = x^8 \), \( \frac{d}{dx} (3x^5) = 15x^4 \), and \( \frac{d}{dx} (-16x) = -16 \). Therefore, ƒ'(x) = x^8 + 15x^4 - 16.
To find the intervals where ƒ is increasing or decreasing, solve ƒ'(x) = 0 to find critical points. Analyze the sign of ƒ'(x) in each interval determined by these critical points.
Next, find the second derivative ƒ''(x) to determine concavity. Differentiate ƒ'(x): \( \frac{d}{dx} (x^8) = 8x^7 \), \( \frac{d}{dx} (15x^4) = 60x^3 \), and \( \frac{d}{dx} (-16) = 0 \). Thus, ƒ''(x) = 8x^7 + 60x^3.
To find intervals of concavity, solve ƒ''(x) = 0 to find inflection points. Analyze the sign of ƒ''(x) in each interval determined by these inflection points to determine where the function is concave up or concave down.

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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 involves analyzing the first derivative of a function, ƒ'(x), to determine where the function is increasing or decreasing. If ƒ'(x) > 0 on an interval, the function is increasing; if ƒ'(x) < 0, it is decreasing. Critical points, where ƒ'(x) = 0 or is undefined, are key to identifying these intervals.
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The First Derivative Test: Finding Local Extrema

Second Derivative Test

The Second Derivative Test uses the second derivative, ƒ''(x), to assess the concavity of a function. If ƒ''(x) > 0, the function is concave up, indicating that the slope of the tangent line is increasing. Conversely, if ƒ''(x) < 0, the function is concave down, suggesting that the slope is decreasing. Points where ƒ''(x) = 0 may indicate inflection points.
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The Second Derivative Test: Finding Local Extrema

Critical Points

Critical points are values of x where the first derivative ƒ'(x) is zero or undefined. These points are essential for determining intervals of increase and decrease, as well as for analyzing concavity. By evaluating the behavior of the function around these points, one can classify them as local maxima, minima, or points of inflection.
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Critical Points
Related Practice
Textbook Question

Does ƒ(x) = (x⁶/2) + (5x⁴/4) - 15x² have any inflection points? If so, identify them.

Textbook Question

Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.


lim_x→ 1 ln x / (4x - x² - 3)

Textbook Question

Use ƒ' and ƒ" to complete parts (a) and (b). 


a. Find the intervals on which f is increasing and the intervals on which it is decreasing.


b. Find the intervals on which f is concave up and the intervals on which it is concave down.


ƒ(x) = x√(x +9)

Textbook Question

Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.


lim_x→ -1 (x⁴ + x³ + 2x + 2) / (x + 1)

Textbook Question

Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.

lim_x→2 (x² - 2x / (x² - 6x + 8) 

Textbook Question

Evaluate lim_x→2 (x³ - 3x² + 2) / (x-2) using l’Hôpital’s Rule and then check your work by evaluating the limit using an appropriate Chapter 2 method.