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

Question

anonymous · Answer

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

anonymous · Answer

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

apples · Answer

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

apples · Answer

Right, $$\sqrt{x} + C$$

apples · Answer

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

anonymous · Answer

cool thanks a lot

anonymous · Answer

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