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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 4.R.14
Chapter 4, Problem 4.R.14

Find the critical points of the following functions on the given intervals. Identify the absolute maximum and absolute minimum values (if they exist).
ƒ(x) = 4x¹⸍² - x⁵⸍² on [0, 4]

Verified step by step guidance
1
To find the critical points of the function \( f(x) = 4x^{3/2} - x^{5/2} \), first compute the derivative \( f'(x) \). Use the power rule: \( \frac{d}{dx}[x^n] = nx^{n-1} \).
Calculate \( f'(x) = \frac{d}{dx}[4x^{3/2}] - \frac{d}{dx}[x^{5/2}] = 6x^{1/2} - \frac{5}{2}x^{3/2} \).
Set \( f'(x) = 0 \) to find the critical points: \( 6x^{1/2} - \frac{5}{2}x^{3/2} = 0 \). Factor out \( x^{1/2} \): \( x^{1/2}(6 - \frac{5}{2}x) = 0 \). Solve for \( x \).
The solutions to \( x^{1/2}(6 - \frac{5}{2}x) = 0 \) are \( x = 0 \) and \( 6 - \frac{5}{2}x = 0 \). Solve \( 6 - \frac{5}{2}x = 0 \) to find \( x = \frac{12}{5} \).
Evaluate \( f(x) \) at the critical points \( x = 0 \), \( x = \frac{12}{5} \), and the endpoints \( x = 0 \) and \( x = 4 \) to determine the absolute maximum and minimum values on the interval \([0, 4]\).

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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 and minima, as they represent potential locations where the function's behavior changes. To find critical points, one must first compute the derivative of the function and solve for the values of x that satisfy the condition.
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Absolute Maximum and Minimum

The absolute maximum and minimum values of a function on a closed interval are the highest and lowest values that the function attains within that interval. To determine these values, one must evaluate the function at its critical points and at the endpoints of the interval. The largest and smallest of these values will represent the absolute maximum and minimum, respectively.
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Evaluating Functions on Intervals

When analyzing functions on a specific interval, it is crucial to evaluate the function at both the endpoints and any critical points found within the interval. This process ensures that all potential maximum and minimum values are considered. The function's behavior can vary significantly across different intervals, making this evaluation vital for accurate conclusions.
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