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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 4.1.67
Chapter 4, Problem 4.1.67

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

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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].

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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.
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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.
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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.
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Related Practice
Textbook Question

{Use of Tech} A family of superexponential functions Let ƒ(x) = (a + x)ˣ , where a > 0.



c. Compute ƒ'. Then graphƒ and ƒ' for a = 0.5, 1, 2, and 3.

Textbook Question

Use the graphs of ƒ' and ƒ" to complete the following steps. <IMAGE>

a. Find the critical points of f and determine where f is increasing and where it is decreasing.

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) cos² x on [0,π]

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Textbook Question

Viewing angles An auditorium with a flat floor has a large screen on one wall. The lower edge of the screen is 3 ft above eye level and the upper edge of the screen is 10 ft above eye level (see figure). How far from the screen should you stand to maximize your viewing angle? <IMAGE>

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) = x+ cos⁻¹x on [-1,1]

Textbook Question

Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant.


ƒ(x) = x³ / 3 - 9x