Ch. 4 - Applications of Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 4.1.21

Chapter 4, Problem 4.1.21

Absolute Extrema on Finite Closed Intervals

In Exercises 21–36, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.

f(x) = (2/3)x − 5, −2 ≤ x ≤ 3

Verified step by step guidance

Step 1: Understand the problem. We need to find the absolute maximum and minimum values of the function f(x) = (2/3)x - 5 on the interval [-2, 3]. Absolute extrema occur at critical points or endpoints of the interval.

Step 2: Evaluate the function at the endpoints of the interval. Calculate f(-2) and f(3) to determine the function values at these points. This will help us identify potential extrema.

Step 3: Since f(x) = (2/3)x - 5 is a linear function, it does not have any critical points within the interval. Critical points are found by setting the derivative equal to zero, but the derivative of a linear function is constant and does not equal zero.

Step 4: Compare the function values at the endpoints. The absolute maximum and minimum will be the largest and smallest values obtained from evaluating the function at x = -2 and x = 3.

Step 5: Graph the function f(x) = (2/3)x - 5 over the interval [-2, 3]. Plot the points (-2, f(-2)) and (3, f(3)) on the graph. These points represent the absolute extrema, and their coordinates should be clearly marked.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

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 a function attains on a given interval. To find these values, evaluate the function at critical points and endpoints of the interval. The absolute maximum is the largest value, and the absolute minimum is the smallest value within the interval.

Critical Points

Critical points are values of x where the derivative of the function is zero or undefined. These points are potential candidates for local extrema. In the context of finding absolute extrema, critical points within the interval, along with endpoints, must be evaluated to determine the function's maximum and minimum values.

Graphing Linear Functions

Graphing linear functions involves plotting a straight line based on the function's slope and y-intercept. For f(x) = (2/3)x - 5, the slope is 2/3, and the y-intercept is -5. On the interval [-2, 3], plot the line and identify the coordinates of the endpoints to determine where the absolute extrema occur.

Recommended video:

07:17

Linearization

Related Practice

Textbook Question

Finding Indefinite Integrals

In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.

∫(sin2x − csc²x)dx

views

Textbook Question

Initial Value Problems

Solve the initial value problems in Exercises 71–90.

dv/dt = (1/2)sec t tan t, v(0) = 1

views

Textbook Question

In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.

______

y = √𝓍² ― 1

171

views

Textbook Question

54. Fermat’s principle in optics Light from a source A is reflected by a plane mirror to a receiver at point B, as shown in the accompanying figure. Show that for the light to obey Fermat’s principle, the angle of incidence must equal the angle of reflection, both measured from the line normal to the reflecting surface. (This result can also be derived without calculus. There is a purely geometric argument, which you may prefer.)

322

views

Textbook Question

Finding Critical Points

In Exercises 41–50, determine all critical points and all domain endpoints for each function.

y = x² − 32√x

200

views

Textbook Question

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.

2. y=x^4/4-2x^2+4

223

views