Find f(x) if f'(x)=(3/x^5) and f(1/2)=1. Please give explanation with steps for best answer. thanks!!?
回答 (3)
2015-08-06 5:42 am
f(x) = (-3/4)x^-4 + C
1 = -12 + C
f(x) = (-3/4)x^-4 + 13
1 = -12 + C
f(x) = (-3/4)x^-4 + 13
2015-08-06 5:31 am
You have to take the antiderivative of f ' (x) and replace x with 1/2 and f(x) with 1
f '(x) = 3/x^5
f(x) = -3/4x^4 + C
f(1/2) = -3/4(1/2)^4 + C
1 = -3/4(1/16) + C
1 = -3/(1/4) + C
1 + 12 = C
13 = C
so f(x) = -3/4x^4 + 13
f '(x) = 3/x^5
f(x) = -3/4x^4 + C
f(1/2) = -3/4(1/2)^4 + C
1 = -3/4(1/16) + C
1 = -3/(1/4) + C
1 + 12 = C
13 = C
so f(x) = -3/4x^4 + 13
參考: That is if I did it right lol
2015-08-06 5:26 am
you can't integrate ? 3 x^(-5) ---> - 3/4 x^-4 + C , where C = 12
收錄日期: 2021-05-01 22:45:06
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