1. ## stationary points

Hello there,

I have a question where i have to find stationary points and then determin whether they are maximum or minimum:
y=2sin(x) -(x)

i get dy/dx = 2cos(x) - 1 and equal to zero i get 1/3 pi. With it obviously being a sine curve, how can i find the next stationary point?

Also d^2y/dx^2 i get:
-2sin(x)
which at 1/3 pi gives me -3^1/2 but in the anser book its minus root 3 minus third pi... where have i gone wrong?

Thanks!

2. Originally Posted by darksupernova
Hello there,

I have a question where i have to find stationary points and then determin whether they are maximum or minimum:
y=2sin(x) -(x)

i get dy/dx = 2cos(x) - 1 and equal to zero i get 1/3 pi. With it obviously being a sine curve, how can i find the next stationary point?

Also d^2y/dx^2 i get:
-2sin(x)
which at 1/3 pi gives me -3^1/2 but in the anser book its minus root 3 minus third pi... where have i gone wrong?

Thanks!

The second derivative also correct.

Notice though, that if you set the derivative equal to zero, that means

$2\cos{x} - 1 = 0$

$2\cos{x} = 1$

$\cos{x} = \frac{1}{2}$

$x = \left\{\frac{\pi}{3}, 2\pi - \frac{\pi}{3}\right\} + 2\pi n$ where $n \in \mathbf{Z}$

$x = \left\{\frac{\pi}{3}, \frac{5\pi}{3}\right\} + 2\pi n$ where $n \in \mathbf{Z}$.

Now you need to do the second derivative test on each of these values. Luckily, the periodicity of the sine function will mean that $\frac{\pi}{3}$ will have the same value as any of its $2\pi n$ multiples, as will $\frac{5\pi}{3}$.

3. ah of course, its just 2pi minus the bit you have because of its symmetry!

Thanks

4. Originally Posted by darksupernova
ah of course, its just 2pi minus the bit you have because of its symmetry!

Thanks
So now, using the second derivative test

$y''\left(\frac{\pi}{3} + 2\pi n\right) = -2\sin{\left(\frac{\pi}{3} + 2\pi n\right)}$

$= -2\left(\frac{\sqrt{3}}{2}\right)$

$= -\sqrt{3}$.

Since the second derivative is negative, that means the stationary points $\frac{\pi}{3} + 2\pi n$ are all maxima.

$y''\left(\frac{5\pi}{3} + 2\pi n\right) = -2\sin{\left(\frac{5\pi}{3} + 2\pi n\right)}$

$= -2\left(-\frac{\sqrt{3}}{2}\right)$

$= \sqrt{3}$.

Since the second derivative is positive, that means the stationary points $\frac{5\pi}{3} + 2\pi n$ are all minima.