$\displaystyle F(x) = sin(\frac{\pi x}{200})-\frac{x^2}{10^4}$

On 0<v<100.

We're supposed to use the First and Second Derivatives to prove that the function has only one local maximum, but we're not supposed to solve for the critical numbers of F(x). We're supposed to look at the concavity.

So first, my two derivatives:

For the 'left':$\displaystyle W(x) = sin(\frac{\pi x}{200})$

$\displaystyle W'(x) = cos(\frac{\pi x}{200})*\frac{200\pi}{200^2}=$$\displaystyle \frac{\pi cos(\frac{\pi x}{200})}{200}$

And then the 'right': $\displaystyle W(x) = \frac{x^2}{10^4}$

$\displaystyle W'(x) = 2x^{-4}$

So $\displaystyle F'(x) = \frac{\pi cos(\frac{\pi x}{200})}{200} - 2x^{-4}$

And then the second derivative:

$\displaystyle W(x) = \frac{\pi cos(\frac{\pi x}{200})}{200}$

$\displaystyle W'(x) = \pi (-sin(\frac{\pi x}{200}))*\frac{\pi}{200^2}$ = $\displaystyle \frac{-\pi ^2sin(\frac{\pi x}{200})}{200^2}$

And then to take the derivative of the whole fraction and subtract the 'right' part:

$\displaystyle F''(x) = -\frac{\pi ^2sin(\frac{\pi x}{200})}{200^4}+8x^{-5}$

Okay, my first question is are the two above correct? And if they are, since I'm only looking at the concavity, would it be valid to find the inflection point in the domain, and since it's concave up from [0,i] and concave down from [i,100], that proves there's only one maximum?

I know this was a bit of a lengthy problem, but thanks in advance for any help.