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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 4.1.89b
Chapter 4, Problem 4.1.89b

Values of related functions Suppose f is differentiable on (-∞,∞) and assume it has a local extreme value at the point x = 2, where f(2) = 0. Let g(x) = xf(x) + 1 and let h(x) = xf(x) + x +1, for all values of x.


b. Does either g or h have a local extreme value at x = 2? Explain.

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To determine if g(x) or h(x) has a local extreme value at x = 2, we need to find their derivatives and evaluate them at x = 2.
First, find the derivative of g(x) = x * f(x) + 1. Use the product rule for differentiation: if u(x) = x and v(x) = f(x), then the derivative g'(x) = u'(x)v(x) + u(x)v'(x).
Calculate g'(x): g'(x) = 1 * f(x) + x * f'(x) = f(x) + x * f'(x).
Evaluate g'(x) at x = 2: g'(2) = f(2) + 2 * f'(2). Since f(2) = 0 and f has a local extreme at x = 2, f'(2) = 0. Thus, g'(2) = 0.
Now, find the derivative of h(x) = x * f(x) + x + 1. Differentiate h(x) using the sum and product rules: h'(x) = f(x) + x * f'(x) + 1. Evaluate h'(x) at x = 2: h'(2) = f(2) + 2 * f'(2) + 1. Since f(2) = 0 and f'(2) = 0, h'(2) = 1, indicating h does not have a local extreme at x = 2.

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Key Concepts

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

Local Extreme Values

A local extreme value of a function occurs at a point where the function reaches a maximum or minimum relative to its immediate surroundings. For a function to have a local extreme value at a point, the derivative at that point must be zero, indicating a horizontal tangent. In this case, since f has a local extreme at x = 2, we know that f'(2) = 0.
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Differentiability and Derivatives

A function is differentiable at a point if it has a defined derivative there, which means it is smooth and has no sharp corners or discontinuities. The derivative provides information about the function's rate of change and is crucial for determining local extreme values. Since f is differentiable everywhere, we can analyze the derivatives of g and h to assess their behavior at x = 2.
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Product Rule in Differentiation

The product rule is a formula used to find the derivative of the product of two functions. If u(x) and v(x) are functions, the derivative of their product is given by u'v + uv'. This rule is essential for differentiating g(x) = xf(x) and h(x) = xf(x) + x + 1, as both functions involve the product of x and f(x), allowing us to determine if they have local extreme values at x = 2.
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Related Practice
Textbook Question

Pen problems


b. A rancher plans to make four identical and adjacent rectangular pens against a barn, each with an area of 100 m² (see figure). What are the dimensions of each pen that minimize the amount of fence that must be used? <IMAGE>

Textbook Question

{Use of Tech} Basketball shot A basketball is shot with an initial velocity of v ft/s at an angle of 45° to the floor. The center of the basketball is 8 ft above the floor at a horizontal distance of 18 feet from the center of the basketball hoop when it is released. The height h (in feet) of the center of the basketball after it has traveled a horizontal distance of x feet is modeled by the function h(x) = 32x² / v² + x + 8 (see figure). <IMAGE>



b. During the flight of the basketball, show that the distance s from the center of the basketball to the front of the hoop is s = √ (x - 17.25)² + ( -(4x² / 81) + x - 2)² (Hint: The diameter of the basketball hoop is 18 inches.) 

Textbook Question

Values of related functions Suppose f is differentiable on (-∞,∞) and assume it has a local extreme value at the point x = 2, where f(2) = 0. Let g(x) = xf(x) + 1 and let h(x) = xf(x) + x +1, for all values of x.


a. Evaluate g(2), h(2), g'(2), and h'(2).

Textbook Question

{Use of Tech} A damped oscillator The displacement of an object as it bounces vertically up and down on a spring is given by y(t) = 2.5e⁻ᵗ cos 2t, where the initial displacement is y(0) = 2.5 and y = 0 corresponds to the rest position (see figure). <IMAGE>

b. Find the time and the displacement when the object reaches its lowest point.

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Textbook Question

{Use of Tech} Every second counts You must get from a point P on the straight shore of a lake to a stranded swimmer who is 50 from a point Q on the shore that is 50 m from you (see figure). Assuming that you can swim at a speed of 2 m/s and run at a speed of 4 m/s, the goal of this exercise is to determine the point along the shore, x meters from Q, where you should stop running and start swimming to reach the swimmer in the minimum time. <IMAGE>


b. Find the critical point of T on (0, 50).

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Textbook Question

{Use of Tech} Let f(x) = ln((x+1)/(x-1)) and g(x) = ln ((x+1)/(x-1)).

b. Sketch graphs of f and g to show that these functions do not differ by a constant.