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anonymous

  • 5 years ago

f'x=1/(2x^1/2) and f(1)=2, find f(1/2)

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  1. anonymous
    • 5 years ago
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    How do I go about solving this? Do I find the antiderivative?

  2. anonymous
    • 5 years ago
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    Is this f'(x)=1/(2(x)^1/2) or f'(x)=1/(2x)^1/2 ?

  3. apples
    • 5 years ago
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    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.

  4. anonymous
    • 5 years ago
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    sorry, it's 1/ ( 2 (x)^1/2)

  5. anonymous
    • 5 years ago
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    I found that the antiderivate is x^1/2, is that what you got

  6. anonymous
    • 5 years ago
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    Your antiderivative is going to be x^(1/2)+c

  7. apples
    • 5 years ago
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    Right, \[\sqrt{x} + C\]

  8. apples
    • 5 years ago
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    \[f(x) = \int \frac{1}{2\sqrt{x}} dx = \sqrt{x} + C\]\[\sqrt{1}+C=2\] Thus C is 1 and \[f(\frac{1}{2}) = \sqrt{\frac{1}{2}} + 1\]

  9. anonymous
    • 5 years ago
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    cool thanks a lot

  10. anonymous
    • 5 years ago
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    anyone wanna help for lim x->infinity x*sin(1/x), is sin(1/x) 0?

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