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 4.1.67
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) = x²/₃(4 -x²) on -2,2

1
First, understand that absolute extrema refer to the highest and lowest values of a function on a given interval. We need to find these values for ƒ(x) = x^(2/3)(4 - x^2) on the interval [-2, 2].
To find the extrema, start by finding the derivative of the function ƒ(x). Use the product rule for differentiation, which states that if you have a function that is the product of two functions, u(x) and v(x), then its derivative is u'(x)v(x) + u(x)v'(x).
Set the derivative equal to zero to find critical points. Critical points are where the derivative is zero or undefined, and they are potential candidates for extrema. Solve the equation to find the values of x that make the derivative zero.
Evaluate the function ƒ(x) at the critical points found in the previous step, as well as at the endpoints of the interval, x = -2 and x = 2. This will give you the values of the function at these points.
Compare the values obtained from evaluating the function at the critical points and endpoints. The largest value will be the absolute maximum, and the smallest value will be the absolute minimum on the interval [-2, 2].

This video solution was recommended by our tutors as helpful for the problem above
Video duration:
6mPlay a video:
Was this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Absolute Extrema
Absolute extrema refer to the highest and lowest values of a function over a specified interval. To find these values, one must evaluate the function at critical points, where the derivative is zero or undefined, as well as at the endpoints of the interval. The largest of these values is the absolute maximum, while the smallest is the absolute minimum.
Recommended video:
Finding Extrema Graphically
Critical Points
Critical points are values of the independent variable where the derivative of the function is either zero or does not exist. These points are essential in determining the behavior of the function, as they can indicate potential locations for local maxima or minima. To find critical points, one must first compute the derivative of the function and solve for when it equals zero or is undefined.
Recommended video:
Critical Points
Endpoints of the Interval
Endpoints of the interval are the boundary values of the domain over which the function is being 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 of the function. For the given interval [-2, 2], the function must be evaluated at both x = -2 and x = 2.
Recommended video:
Estimating the Area Under a Curve with Right Endpoints & Midpoint
Related Videos
Related Practice