Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals3h 25m
5. Graphical Applications of Derivatives
Finding Global Extrema
Problem 4.R.14
Textbook Question
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]

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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