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 Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
5. Graphical Applications of Derivatives
Finding Global Extrema
Problem 4.R.8
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).
g(x) = x⁴ - 50x² on [-1, 5]

1
To find the critical points of the function \( g(x) = x^4 - 50x^2 \), first compute the derivative \( g'(x) \).
The derivative is \( g'(x) = 4x^3 - 100x \). Set \( g'(x) = 0 \) to find the critical points.
Solve the equation \( 4x^3 - 100x = 0 \). Factor out the common term: \( 4x(x^2 - 25) = 0 \).
This gives \( x = 0 \) or \( x^2 - 25 = 0 \). Solving \( x^2 - 25 = 0 \) gives \( x = 5 \) or \( x = -5 \).
Evaluate \( g(x) \) at the critical points \( x = 0, 5, -5 \) and at the endpoints of the interval \( x = -1 \) and \( x = 5 \) to determine the absolute maximum and minimum values.

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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 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 be the absolute maximum and minimum, respectively.
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Endpoints of an Interval
Endpoints of an interval are the boundary values that define the limits of the interval in which a function is analyzed. In the context of finding absolute extrema, it is crucial to evaluate the function at these endpoints, as they can potentially yield the highest or lowest values alongside the critical points. For the function g(x) = x⁴ - 50x² on [-1, 5], the endpoints are x = -1 and x = 5.
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