Skip to main content
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 4.1.43
Chapter 4, Problem 4.1.43

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 - 10 on [-2, 3]

Verified step by step guidance
1
First, understand that absolute extrema refer to the highest and lowest values a function can take on a given interval. We need to find these values for the function ƒ(x) = x² - 10 on the interval [-2, 3].
To find the absolute extrema, we need to evaluate the function at critical points and endpoints. Start by finding the derivative of ƒ(x). The derivative, ƒ'(x), is obtained by differentiating ƒ(x) = x² - 10, which gives ƒ'(x) = 2x.
Set the derivative equal to zero to find critical points: 2x = 0. Solving this equation gives x = 0. This is a critical point within the interval [-2, 3].
Evaluate the function ƒ(x) at the critical point and at the endpoints of the interval. Calculate ƒ(-2), ƒ(0), and ƒ(3). These values will help determine the absolute maximum and minimum.
Compare the values obtained from ƒ(-2), ƒ(0), and ƒ(3). The largest value will be the absolute maximum, and the smallest value will be the absolute minimum on the interval [-2, 3].

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m
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:
05:58
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. In the context of finding absolute extrema, critical points must be evaluated alongside the endpoints of the interval.
Recommended video:
04:50
Critical Points

Closed Interval

A closed interval, denoted as [a, b], includes all numbers between a and b, including the endpoints a and b themselves. In calculus, analyzing functions over closed intervals is crucial for finding absolute extrema, as it ensures that both the endpoints and any critical points within the interval are considered. This guarantees a comprehensive evaluation of the function's behavior across the entire interval.
Recommended video:
05:12
Finding Global Extrema (Extreme Value Theorem)
Related Practice
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) = sin 3x on [-π/4,π/3]

Textbook Question

Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.

lim_x→∞ (log₂ x - log₃ x)

Textbook Question

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.


lim_x→ 0⁺ (x - 3 √x) / (x - √x)

Textbook Question

Differentials Consider the following functions and express the relationship between a small change in x and the corresponding change in y in the form dy = f'(x)dx.


f(x) = 3x³ - 4x

Textbook Question

Finding antiderivatives. Find all the antiderivatives of the following functions. Check your work by taking derivatives.


ƒ(x) = eˣ

Textbook Question

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.


lim_x→0 csc 6x sin 7x