Find the coordinates of the first inflection point to the right of the origin:

$\displaystyle Si(x) = \int^x_0 \frac{sin t}{t}dt$

I think that it's obvious that I have to find the second derivative of Si(x) and then set it equal to zero. Here is what I did:

Let $\displaystyle g(x) = Si(x) = \int^x_0 \frac{sin t}{t}dt$

$\displaystyle g'(x) = \frac{dg}{dx} = \frac{d}{dx}[\int^x_0 \frac{sin t}{t}dt = \frac{sin(x)}{x}$ (according to FTC1)

$\displaystyle g''(x) = (\frac{sin(x)}{x})' = (\sin(x)(x^{-1}))' = \cos(x)(x^{-1}) + \sin(x)(-x^{-2}) = \frac{x \cos(x) - \sin(x)}{x^2}$

Assuming all is well to here (big assumption, I know), I get stuck on solving for zero:

$\displaystyle 0 = x \cos(x) - \sin(x)$

When I graph this on my calculator, there isn't an obvious solution.

For whatever it's worth, my calculator tells me that the answer to this is approximately $\displaystyle \frac{2}{3}\pi$

Anybody see where I am going wrong or have a suggestion to get me unstuck?

Thanks.