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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 4, Problem 4.4.2

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

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To find the inflection points, local maxima, and minima, start by finding the first derivative of the function y = \(\frac{x^4}{4}\) - 2x^2 + 4. The first derivative, y', will help identify critical points where the slope is zero or undefined.
Calculate the first derivative: y' = \(\frac{d}{dx}\)(\(\frac{x^4}{4}\) - 2x^2 + 4). This simplifies to y' = x^3 - 4x.
Set the first derivative equal to zero to find critical points: x^3 - 4x = 0. Factor the equation: x(x^2 - 4) = 0, which gives x = 0, x = 2, and x = -2.
To determine if these critical points are local maxima, minima, or points of inflection, calculate the second derivative: y'' = \(\frac{d}{dx}\)(x^3 - 4x). This simplifies to y'' = 3x^2 - 4.
Evaluate the second derivative at the critical points: y''(0), y''(2), and y''(-2). If y'' > 0, the function is concave up (local minima); if y'' < 0, the function is concave down (local maxima); if y'' = 0, it may be an inflection point. Analyze the sign changes in y'' to identify intervals of concavity.

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Key Concepts

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

Inflection Points

Inflection points are points on a curve where the concavity changes, meaning the curve transitions from being concave up to concave down or vice versa. To find inflection points, one must analyze the second derivative of the function. If the second derivative changes sign at a point, that point is an inflection point.
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Critical Points

Local Maxima and Minima

Local maxima and minima are points where a function reaches a peak or a trough within a certain interval. These points can be identified using the first derivative test, where critical points (where the first derivative is zero or undefined) are evaluated to determine if they correspond to a maximum, minimum, or neither based on the sign of the derivative before and after the point.
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The First Derivative Test: Finding Local Extrema

Concavity and Differentiability

Concavity refers to the direction of the curvature of a function, determined by the sign of the second derivative. A function is concave up when its second derivative is positive and concave down when it is negative. Differentiability indicates that a function has a derivative at all points in an interval, which is essential for determining local extrema and concavity.
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Determining Concavity Given a Function
Related Practice
Textbook Question

Identifying Extrema


In Exercises 61 and 62, the graph of f' is given. Assume that f is continuous, and determine the x-values corresponding to local minima and local maxima.


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Textbook Question

Identifying Extrema


In Exercises 61 and 62, the graph of f' is given. Assume that f is continuous, and determine the x-values corresponding to local minima and local maxima.


178
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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

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Textbook Question

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

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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

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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.)

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