Question

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

Answer 1

How do I go about solving this? Do I find the antiderivative?

Answer 2

Is this f'(x)=1/(2(x)^1/2) or f'(x)=1/(2x)^1/2 ?

Answer 3

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.

Answer 4

sorry, it's 1/ ( 2 (x)^1/2)

Answer 5

I found that the antiderivate is x^1/2, is that what you got

Answer 6

Your antiderivative is going to be x^(1/2)+c

Answer 7

Right, \[\sqrt{x} + C\]

Answer 8

\[f(x) = \int \frac{1}{2\sqrt{x}} dx = \sqrt{x} + C\]\[\sqrt{1}+C=2\]
Thus C is 1 and \[f(\frac{1}{2}) = \sqrt{\frac{1}{2}} + 1\]

Answer 9

cool thanks a lot

Answer 10

anyone wanna help for lim x->infinity x*sin(1/x), is sin(1/x) 0?