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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.3.53a
Chapter 4, Problem 4.3.53a

Identifying Extrema


In Exercises 53–60:


a. Find the local extrema of each function on the given interval, and say where they occur.


f(x) = sin 2x, 0 ≤ x ≤ π

Verified step by step guidance
1
To find the local extrema of the function f(x) = sin(2x) on the interval [0, π], we first need to find the critical points. This involves taking the derivative of the function and setting it equal to zero.
The derivative of f(x) = sin(2x) is f'(x) = 2cos(2x). Set this equal to zero to find the critical points: 2cos(2x) = 0.
Solve the equation cos(2x) = 0. The solutions to this equation are 2x = π/2 + kπ, where k is an integer. Divide by 2 to solve for x: x = π/4 + kπ/2.
Determine which of these critical points lie within the interval [0, π]. For k = 0, x = π/4; for k = 1, x = 3π/4. Both are within the interval.
Evaluate the function f(x) at the critical points and the endpoints of the interval, x = 0 and x = π, to determine the local extrema. Compare these values to identify the local maxima and minima.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Local Extrema

Local extrema refer to the points in a function where it reaches a local maximum or minimum value. These points occur where the derivative of the function is zero or undefined, indicating a change in the direction of the function's slope. Identifying local extrema involves finding these critical points and evaluating the function's behavior around them.
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Finding Extrema Graphically

Derivative and Critical Points

The derivative of a function provides the rate of change or slope of the function at any given point. Critical points occur where the derivative is zero or undefined, which are potential locations for local extrema. To find these points, differentiate the function and solve for the values of x that satisfy these conditions within the given interval.
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Critical Points

Trigonometric Functions and Their Derivatives

Understanding the behavior of trigonometric functions like sine is crucial for solving problems involving them. The derivative of sin(2x) is 2cos(2x), which helps in finding critical points. Analyzing the sine and cosine functions over the interval [0, π] allows us to determine where the function increases or decreases, aiding in identifying local extrema.
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Introduction to Trigonometric Functions
Related Practice
Textbook Question

53. Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yd) and continued to do so all day. Ship B was sailing east at 8 knots and continued to do so all day.

a. Start counting time with t=0 at noon and express the distance s between the ships as a function of t.

Textbook Question

Finding Antiderivatives

In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.

−3x⁻⁴

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Textbook Question

Identifying Extrema


In Exercises 53–60:


a. Find the local extrema of each function on the given interval, and say where they occur.


f(x) = x / 2 − 2sin (x/2), 0 ≤ x ≤ 2π

Textbook Question

Identifying Extrema


In Exercises 41–52:


a. Identify the function’s local extreme values in the given domain, and say where they occur.


f(t) = 12t − t³, −3 ≤ t < ∞

Textbook Question

Identifying Extrema


In Exercises 53–60:


a. Find the local extrema of each function on the given interval, and say where they occur.


f(x) = csc²x − 2cot x, 0 < x < π

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Textbook Question

25. Paper folding A rectangular sheet of 8.5-in.-by-11-in. paper is placed on a flat surface. One of the corners is placed on the opposite longer edge, as shown in the figure, and held there as the paper is smoothed flat. The problem is to make the length of the crease as small as possible. Call the length L. Try it with paper.

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a. Show that L^2=2x^3/(2x-8.5).