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Find the minimum and maximum values of the function on the given interval by comparing values at the critical points and endpoints. f(x) = 5 tan¹ (x) - x. [1.9] (Express numbers in exact form. Use symbolic notation and fractions where needed.) minimum: 3.5357436 Incorrect aximum: incorrect -2

Find the minimum and maximum values of the function on the given interval by comparing values at the critical points and endpoints. f(x) = 5 tan¹ (x) - x. [1.9] (Express numbers in exact form. Use symbolic notation and fractions where needed.) minimum: 3.5357436 Incorrect aximum: incorrect -2

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Your answer is wrong They said write anwer in exact form so plz give me answer in exact form not in points Are u really mad I am not mad that posting again and again Plz give me exactly soloution
Find the minimum and maximum values of the function on the given interval by comparing values at the critical points and endpoints. f(x) = 5 tan¹ (x) - x. [1,9] (Express numbers in exact form. Use symbolic notation and fractions where needed.) minimum: 3.5357436 Incorrect maximum: incorrect
Transcribed Image Text:Find the minimum and maximum values of the function on the given interval by comparing values at the critical points and endpoints. f(x) = 5 tan¹ (x) - x. [1,9] (Express numbers in exact form. Use symbolic notation and fractions where needed.) minimum: 3.5357436 Incorrect maximum: incorrect
F f(1) check valve x = 2 -2 are 2 = ‡ (9) x² = 4 = = #2 (2) = 5 tam ¹2 -2 5 tom²¹1 - 1 of function 5 tam (2) = 5 ton (2) +2 Maximum value Minimum value 9-9 critical point = 0.78539816 X5 -1 = 2.9269908 = 5x1.10714872 -2 = 3.5357436 fi at critical points and end boints. - 5X -|.10714872 t2 = -3.535746 = 5x1.46013911-9 =-1.69930445 3.5357436 -3.535 74 36 2 at se=2 at se=-2 Was this solution helpful? √x 8
Transcribed Image Text:F f(1) check valve x = 2 -2 are 2 = ‡ (9) x² = 4 = = #2 (2) = 5 tam ¹2 -2 5 tom²¹1 - 1 of function 5 tam (2) = 5 ton (2) +2 Maximum value Minimum value 9-9 critical point = 0.78539816 X5 -1 = 2.9269908 = 5x1.10714872 -2 = 3.5357436 fi at critical points and end boints. - 5X -|.10714872 t2 = -3.535746 = 5x1.46013911-9 =-1.69930445 3.5357436 -3.535 74 36 2 at se=2 at se=-2 Was this solution helpful? √x 8
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