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(3) Consider f(x) = 2x³ + x² - 4x + 1 (a) Find decreasing and increasing intervals of f(x) f'(x) = 6x² + 5x -4 Need critical points ? f'w=0 6x² +5x-4-0 (factor expression) (3x²+ 2x+x-2)=0 <(+2)(-+)² No + correct 3042:0 3x =-2 12x 75 = 0 x=1 K (b) Find concave up and concave down intervals of f(x) Concavily you use the seconde derivative f"(x) = 6x² + 5x-4 = 12x +5 12x=-5 x = -5 Increasing: (-0, $) U (1,00) factoring (Decreasing: (1,5) ,5) Concave up when F) 200 Concave down when FW 20 € (0,0) (c) Find the x-coordinate of any inflection points. snflection rainko one az ? fex) | points (+) (+) ه عله بر 5/8 2 % come from 2nd derivative %2

(3) Consider f(x) = 2x³ + x² - 4x + 1 (a) Find decreasing and increasing intervals of f(x) f'(x) = 6x² + 5x -4 Need critical points ? f'w=0 6x² +5x-4-0 (factor expression) (3x²+ 2x+x-2)=0 <(+2)(-+)² No + correct 3042:0 3x =-2 12x 75 = 0 x=1 K (b) Find concave up and concave down intervals of f(x) Concavily you use the seconde derivative f"(x) = 6x² + 5x-4 = 12x +5 12x=-5 x = -5 Increasing: (-0, $) U (1,00) factoring (Decreasing: (1,5) ,5) Concave up when F) 200 Concave down when FW 20 € (0,0) (c) Find the x-coordinate of any inflection points. snflection rainko one az ? fex) | points (+) (+) ه عله بر 5/8 2 % come from 2nd derivative %2

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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6 (d) Find any relative maxima and minima using the first derivative test. (Justify your answer) ficx) 6x² +5x-4 6x² +5x -4 = 0 2 (3x² - 2x-xx-2)=0 <^^x)=0) ? 2 (3x+2) =0 Relative max 0 112 value of 30 x=2 -> How? zero ? No relative value since everything is. f"(x) = 12x + 5 12x +5=0 12x = -5 (e) Find any relative maxima and minima using the second derivative test. (Justify your answer) A aweb wwabnos bas qurosond x = = 15/2 12 (1)+5=1 No relative it equals to zove тах со Relative min value of o @ x = 15/2 ਕਰਦੇ 1/6 How 7 DAW7 1½ 2/12
Transcribed Image Text:6 (d) Find any relative maxima and minima using the first derivative test. (Justify your answer) ficx) 6x² +5x-4 6x² +5x -4 = 0 2 (3x² - 2x-xx-2)=0 <^^x)=0) ? 2 (3x+2) =0 Relative max 0 112 value of 30 x=2 -> How? zero ? No relative value since everything is. f"(x) = 12x + 5 12x +5=0 12x = -5 (e) Find any relative maxima and minima using the second derivative test. (Justify your answer) A aweb wwabnos bas qurosond x = = 15/2 12 (1)+5=1 No relative it equals to zove тах со Relative min value of o @ x = 15/2 ਕਰਦੇ 1/6 How 7 DAW7 1½ 2/12
(3) Consider f(x) = 2x³ + x² - 4x + 1 (a) Find decreasing and increasing intervals of f(x) f'(x) = Need critical points f'(x=0 6x²+5x-4 = 0 (factor expression) Not (3 (³x² + 2x + x - 2) = 0 < (23)+2) (x - 1)2 Not corred correct factoring 3xx72=0 X=1 ? کے دو = 6x² + 5x - 4 3x =-2 = a/n f"(x) = 6x² + 5x-4 = 12x +5 12x +5 = 0 дл (b) Find concave up and concave down intervals of f(x) Concavily you use the second derivative 12x=05 12 x = 3/2 Increasing: (-00, $) U (1,0) Decreasing: دنا Concave up Concave down when FW 20 $1 when FW) 20 (-∞, 32) F(x) too, 9) IS (c) Find the x-coordinate of any inflection points, Inflection points are وال Als be sale of 5/8 als "C points (+) (0% 2 = come from 2nd derivative 6/ 9/2
Transcribed Image Text:(3) Consider f(x) = 2x³ + x² - 4x + 1 (a) Find decreasing and increasing intervals of f(x) f'(x) = Need critical points f'(x=0 6x²+5x-4 = 0 (factor expression) Not (3 (³x² + 2x + x - 2) = 0 < (23)+2) (x - 1)2 Not corred correct factoring 3xx72=0 X=1 ? کے دو = 6x² + 5x - 4 3x =-2 = a/n f"(x) = 6x² + 5x-4 = 12x +5 12x +5 = 0 дл (b) Find concave up and concave down intervals of f(x) Concavily you use the second derivative 12x=05 12 x = 3/2 Increasing: (-00, $) U (1,0) Decreasing: دنا Concave up Concave down when FW 20 $1 when FW) 20 (-∞, 32) F(x) too, 9) IS (c) Find the x-coordinate of any inflection points, Inflection points are وال Als be sale of 5/8 als "C points (+) (0% 2 = come from 2nd derivative 6/ 9/2
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