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Author Question: Find the particular solution determined by the given condition.f'(x) = x2/5+ x; f(1) = -6 (Read 55 times)

Kthamas

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on: Jun 16, 2021

Question 1

Find the particular solution determined by the given condition.

f'(x) = 6x2 - 4x + 22;  f(1) = 19
◦ f(x) = 6x3 - 4x2 + 22x - 5
◦ f(x) = 2x3 - 2x2 + 22x + 3
◦ f(x) = 2x3 - 2x2 + 22x - 3
◦ f(x) = 2x3 - 4x2 + 22x - 1

Question 2

Find the particular solution determined by the given condition.

f'(x) = x2/5 + x;  f(1) = -6
◦ f(x) = x1/5 + x2 - 
◦ f(x) = x7/5 + x2 - 
◦ f(x) = x7/5 - x2 + 
◦ f(x) = x7/5 + 2x2 + 
Calculus and Its Applications ¦ Bittinger, Ellenbogen, Surgent ¦ 11th Edition
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Marked as best answer by Kthamas on Jun 16, 2021

reelove4eva

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Reply #1 on: Jun 16, 2021
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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Kthamas

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Reply 2 on: Jun 16, 2021
:D TYSM

billybob123

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Reply 3 on: Yesterday
Great answer, keep it coming :)

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