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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 20
Chapter 4, Problem 20

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)

Verified step by step guidance
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Step 1: Find the first derivative ƒ'(x) of the function ƒ(x) = x√(x + 9). Use the product rule and chain rule to differentiate. The product rule states that if you have two functions u(x) and v(x), then the derivative of their product is u'(x)v(x) + u(x)v'(x). Here, let u(x) = x and v(x) = √(x + 9).
Step 2: Simplify the expression for ƒ'(x) and set it equal to zero to find the critical points. These critical points will help determine where the function is increasing or decreasing. Solve the equation ƒ'(x) = 0 for x.
Step 3: Use the critical points to test intervals on the number line. Choose test points in each interval and substitute them into ƒ'(x) to determine the sign of the derivative. If ƒ'(x) > 0, the function is increasing on that interval; if ƒ'(x) < 0, the function is decreasing.
Step 4: Find the second derivative ƒ''(x) to analyze the concavity of the function. Differentiate ƒ'(x) to obtain ƒ''(x). This will involve applying the product rule and chain rule again.
Step 5: Determine the intervals of concavity by setting ƒ''(x) equal to zero and solving for x to find possible inflection points. Test intervals around these points using test values to determine the sign of ƒ''(x). If ƒ''(x) > 0, the function is concave up on that interval; if ƒ''(x) < 0, the function is 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 of a function, denoted as ƒ', provides information about the function's increasing and decreasing behavior. If ƒ' > 0 on an interval, the function is increasing; if ƒ' < 0, it is decreasing. By finding critical points where ƒ' = 0 or is undefined, we can determine the intervals of increase and decrease.
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The First Derivative Test: Finding Local Extrema

Second Derivative Test

The second derivative of a function, denoted as ƒ'', indicates the concavity of the function. If ƒ'' > 0, the function is concave up, suggesting that the slope of the tangent line is increasing. Conversely, if ƒ'' < 0, the function is concave down, indicating that the slope is decreasing. Analyzing points where ƒ'' = 0 helps identify inflection points.
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The Second Derivative Test: Finding Local Extrema

Critical Points

Critical points occur where the first derivative is zero or undefined, and they are essential for determining intervals of increase and decrease. These points can indicate local maxima or minima. Additionally, critical points are also relevant for the second derivative test, as they may correspond to changes in concavity, helping to identify inflection points.
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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→∞ (3x⁴ - x²) / (6x⁴ + 12) 

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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) = x⁹/9 + 3x⁵ - 16x


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)