Thread: Second Derivatives + Concavity approach

1. Second Derivatives + Concavity approach

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

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.

2. for $\displaystyle f(x) = sinx$ the derivative is $\displaystyle f'(x) = cosx * dx/dx = cosx$

following that logic, derivative of $\displaystyle W(x) = sin(\frac{\pi x}{200})$ is $\displaystyle W'(x) = cos(\frac{\pi x}{200}) * \pi/200$ and $\displaystyle W''(x) = - sin(\frac{\pi x}{200}) *\pi/200 * \pi/200$

for $\displaystyle g(x) = x^n$ the derivative is given as $\displaystyle g'(x) = n*x^(n-1)$

so for $\displaystyle M(x) = \frac{x^2}{10^4}$ the derivative is $\displaystyle M'(x) = 2x/10^4$ and the double derivative is $\displaystyle M''(x)2/10^4$

then $\displaystyle F''(x) = W''(x) - M''(x)$

find the inflection points by setting F''(x) to zero and solving for x.

check where it is positive and where it is negative i.e. find concavity

you should end up with one concave down which means there is only one maximum.

how many concave up you end up isn't important here.

3. Where did I go wrong with the sine/cosine part? I got 200^4, not squared, and my work looked valid.

Because it's the quotient rule, shouldn't it be the bottom squared?

4. not sure what you mean. Remember 200^4 or any number for that matter is a constant and can be taken out of the derivative.

5. No, I"m talking about $\displaystyle sin(\frac{\pi x}{200}$. Chain rule is f'(g(x))g'(x).

If f(x) is sin(x), and g(x) is $\displaystyle \frac{\pi x}{200}$, then doesn't g'(x) involve the quotient rule?

$\displaystyle \frac{(derivativeOfTheTop)(bottom)-(derivativeOfBottom)(top)}{bottom^2}$

That would give you $\displaystyle \frac{\pi}{200^2}$, not $\displaystyle \frac{\pi}{200}$.

Am I mistaken?

6. You're not wrong with the derivative part, but you are with the calculation -- if you use the quotient rule there, you get $\displaystyle g'(x) = \frac{ \pi \cdot 200 - 0 \cdot \pi x}{200^2} = \frac{ \pi}{200}$
It can be done alot easier, though -- since $\displaystyle \frac{ \pi}{200}$ is a constant, $\displaystyle g'(x) = \frac{ \pi}{200}$, thus $\displaystyle f'(x) =$...

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

Using that as my antiderivative doesn't make sense, because it's always concave up.

Am I making a syntactical error or do I just not understand the proof itself, because if the function is always concave up doesn't that mean that there are no max/mins on the domain?