anonymous
  • anonymous
f'x=1/(2x^1/2) and f(1)=2, find f(1/2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
How do I go about solving this? Do I find the antiderivative?
anonymous
  • anonymous
Is this f'(x)=1/(2(x)^1/2) or f'(x)=1/(2x)^1/2 ?
apples
  • apples
Yes, then you find the constant of integration by setting your antiderivative equal to 2 and evaluating where x = 1. Once you know the constant, you can evaluate f(1/2) like normal.

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anonymous
  • anonymous
sorry, it's 1/ ( 2 (x)^1/2)
anonymous
  • anonymous
I found that the antiderivate is x^1/2, is that what you got
anonymous
  • anonymous
Your antiderivative is going to be x^(1/2)+c
apples
  • apples
Right, \[\sqrt{x} + C\]
apples
  • apples
\[f(x) = \int \frac{1}{2\sqrt{x}} dx = \sqrt{x} + C\]\[\sqrt{1}+C=2\] Thus C is 1 and \[f(\frac{1}{2}) = \sqrt{\frac{1}{2}} + 1\]
anonymous
  • anonymous
cool thanks a lot
anonymous
  • anonymous
anyone wanna help for lim x->infinity x*sin(1/x), is sin(1/x) 0?

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