## anonymous 5 years ago f'x=1/(2x^1/2) and f(1)=2, find f(1/2)

1. anonymous

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

2. anonymous

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

3. 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.

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. apples

Right, $\sqrt{x} + C$

8. apples

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

9. anonymous

cool thanks a lot

10. anonymous

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