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 4.1.65a
Textbook Question
Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.
ƒ(x) = x³e⁻ˣ on [-1,5]

1
First, understand that absolute maxima and minima refer to the highest and lowest points on the graph of a function within a given interval. To find these, we need to evaluate the function at critical points and endpoints of the interval.
To find critical points, we need to take the derivative of the function ƒ(x) = x³e⁻ˣ. Use the product rule for differentiation: if u(x) = x³ and v(x) = e⁻ˣ, then ƒ'(x) = u'(x)v(x) + u(x)v'(x).
Calculate the derivative: u'(x) = 3x² and v'(x) = -e⁻ˣ. Therefore, ƒ'(x) = 3x²e⁻ˣ - x³e⁻ˣ.
Set the derivative ƒ'(x) = 0 to find critical points. This simplifies to x²(3 - x)e⁻ˣ = 0. Solve for x to find the critical points within the interval [-1, 5].
Evaluate the function ƒ(x) at the critical points found and at the endpoints x = -1 and x = 5. Compare these values to determine the absolute maximum and minimum values on the interval.

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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 are values of x in the domain of a function where the derivative is either zero or undefined. These points are essential for finding absolute maxima and minima, as they indicate where the function's slope changes, potentially leading to extreme values. To locate critical points, one must first compute the derivative of the function and solve for x.
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Endpoints of the Interval
When determining absolute extrema on a closed interval, it is crucial to evaluate the function at both the critical points and the endpoints of the interval. The absolute maximum or minimum could occur at any of these locations. In this case, the interval is [-1, 5], so the function must be evaluated at x = -1 and x = 5 in addition to any critical points found.
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First Derivative Test
The First Derivative Test is a method used to determine whether a critical point is a local maximum, local minimum, or neither. By analyzing the sign of the derivative before and after the critical point, one can infer the behavior of the function. If the derivative changes from positive to negative, the critical point is a local maximum; if it changes from negative to positive, it is a local minimum.
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The First Derivative Test: Finding Local Extrema
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