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The Second Derivative Test cannot be used to conclude that x = 1 is the location of a relative minimum or relative maximum for which of the following functions? f (x) = cos (x? – 1), where f' (x) = –2æ sin(x² – 1) and f" (æ) = –2 sin(x² – 1) – 4x² cos (x² – 1) - B f (x) = e(=-1)°, where f' (æ) = 2 (x – 1) e(z-1)° and f" (æ) = 4(x – 1)°e(=-1)* + 2e(z-1)² f (x) = + x² – 3x + 1, where f' (x) x2 + 2x – 3 and f" (x) = 2x + 2 3 f (x) = x* – 4x³ + 6x? – 4x+1, where f' (x)= 4x³ – 12x² + 12x – 4 and f" (x) = 12x? – 24x + 12

The Second Derivative Test cannot be used to conclude that x = 1 is the location of a relative minimum or relative maximum for which of the following functions? f (x) = cos (x? – 1), where f' (x) = –2æ sin(x² – 1) and f" (æ) = –2 sin(x² – 1) – 4x² cos (x² – 1) - B f (x) = e(=-1)°, where f' (æ) = 2 (x – 1) e(z-1)° and f" (æ) = 4(x – 1)°e(=-1)* + 2e(z-1)² f (x) = + x² – 3x + 1, where f' (x) x2 + 2x – 3 and f" (x) = 2x + 2 3 f (x) = x* – 4x³ + 6x? – 4x+1, where f' (x)= 4x³ – 12x² + 12x – 4 and f" (x) = 12x? – 24x + 12

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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The Second Derivative Test cannot be used to conclude that x = 1 is the location of a relative minimum or relative maximum for which of the following functions? f (x) = (x² – 1), where f' (x) -2æ sin (x? – 1) and f" (x) = –2 sin(x² – 1) – 4x² cos(x² – 1) = COS B f (x) = e(¤-1)°, , where f' (x) = 2 (x – 1) e(¤-1)' and f" (x) = 4(x – 1)²e(=-1)* + 2e(x–1)? f (æ) = + x? 3 3x + 1, where f' (x) = x² + 2x - 3 and f" (x) = 2x + 2 f (x) = x4 – 4x³ + 6x² – 4x + 1, where f' (x) = 4x³ – 12x² + 12x – 4 and f" (x) = 12x² – 24x + 12
Transcribed Image Text:The Second Derivative Test cannot be used to conclude that x = 1 is the location of a relative minimum or relative maximum for which of the following functions? f (x) = (x² – 1), where f' (x) -2æ sin (x? – 1) and f" (x) = –2 sin(x² – 1) – 4x² cos(x² – 1) = COS B f (x) = e(¤-1)°, , where f' (x) = 2 (x – 1) e(¤-1)' and f" (x) = 4(x – 1)²e(=-1)* + 2e(x–1)? f (æ) = + x? 3 3x + 1, where f' (x) = x² + 2x - 3 and f" (x) = 2x + 2 f (x) = x4 – 4x³ + 6x² – 4x + 1, where f' (x) = 4x³ – 12x² + 12x – 4 and f" (x) = 12x² – 24x + 12
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