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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
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Tutti i libri di testoBriggs 3rd EditionCh. 5 - IntegrationProblema 5.R.113c
Capitolo 5, Problema 5.R.113c

Function defined by an integral Let ƒ(𝓍) = ∫₀ˣ (t ― 1)¹⁵ (t―2)⁹ dt .
(c) For what values of 𝓍 does ƒ have local minima? Local maxima?

Guida verificata passo dopo passo
1
Step 1: Recall that to find local minima and maxima of a function, we need to analyze its derivative. The Fundamental Theorem of Calculus tells us that the derivative of ƒ(𝓍) = ∫₀ˣ g(t) dt is ƒ'(𝓍) = g(𝓍). Here, g(t) = (t - 1)¹⁵ (t - 2)⁹.
Step 2: Set ƒ'(𝓍) = g(𝓍) = (𝓍 - 1)¹⁵ (𝓍 - 2)⁹ equal to zero to find critical points. This equation is satisfied when either (𝓍 - 1) = 0 or (𝓍 - 2) = 0. Thus, the critical points are 𝓍 = 1 and 𝓍 = 2.
Step 3: To determine whether these critical points correspond to local minima or maxima, analyze the sign changes of ƒ'(𝓍) = (𝓍 - 1)¹⁵ (𝓍 - 2)⁹ around the critical points. Consider intervals around 𝓍 = 1 and 𝓍 = 2, such as (0, 1), (1, 2), and (2, ∞).
Step 4: Evaluate the behavior of ƒ'(𝓍) in each interval. For example, in the interval (0, 1), both (𝓍 - 1)¹⁵ and (𝓍 - 2)⁹ are negative, making ƒ'(𝓍) positive. In the interval (1, 2), (𝓍 - 1)¹⁵ is positive and (𝓍 - 2)⁹ is negative, making ƒ'(𝓍) negative. In the interval (2, ∞), both terms are positive, making ƒ'(𝓍) positive.
Step 5: Based on the sign changes of ƒ'(𝓍), conclude that 𝓍 = 1 is a local maximum (since ƒ'(𝓍) changes from positive to negative) and 𝓍 = 2 is a local minimum (since ƒ'(𝓍) changes from negative to positive).

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Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation and integration, stating that if a function is defined as an integral, its derivative can be found by evaluating the integrand at the upper limit of integration. This theorem is essential for analyzing the behavior of the function ƒ(𝓍) in the given question.
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Percorso guidato
06:11
Fundamental Theorem of Calculus Part 1

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These points are crucial for determining local maxima and minima, as they represent potential locations where the function's behavior changes. In the context of ƒ(𝓍), finding where the derivative equals zero will help identify these critical points.
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Critical Points

Second Derivative Test

The Second Derivative Test is a method used to classify critical points as local minima, local maxima, or saddle points. By evaluating the second derivative at a critical point, one can determine the concavity of the function at that point. If the second derivative is positive, the point is a local minimum; if negative, it is a local maximum.
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The Second Derivative Test: Finding Local Extrema
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