1. Local Maximum and Minimum

I'm trying to find the local maximum and minimum values for the equation $\displaystyle f(x)=sin2x-x, 0 < x < \pi$ but I seem to keep coming up with the wrong answer.

My working goes like this (in radians):
$\displaystyle f'(x)=2cos2x-1$
$\displaystyle f''(x)=-4sin2x$

Local maximums and minimums appear where the gradient is 0, so:

$\displaystyle 2cos2x-1=0$
$\displaystyle 2cos2x=1$
$\displaystyle cos2x=\frac{1}{2}$
$\displaystyle x=\frac{1}{6}\pi$

$\displaystyle f''(\frac{1}{6}\pi ) = -4sin\frac{1}{3}\pi = -4\times 0.866 = -3.464$

Therefore there is a local maximum at $\displaystyle x=\frac{1}{6}\pi$

What I can't work out is how to find the local minimum in the same range. I think that, because local maximums and minimums are $\displaystyle \pi$ radians apart on the Cosine graph, they will be $\displaystyle x=\frac{1}{2}\pi$ radians apart on the graph of Cos 2x, which would put the minimum point at $\displaystyle x=\frac{2}{3}\pi$ but the book I'm working from puts it at $\displaystyle x=\frac{5}{6}\pi$.

2. Re: Local Maximum and Minimum

You have determined the $\displaystyle x$-coordinate of the maximum correctly, but notice your equation has two solutions ...

3. Re: Local Maximum and Minimum

Thanks, I can see there's a second solution, but I make it 2/3 pi while the book I'm working from makes it 5/6 pi and I'm not sure why I'm arriving at a different answer.

4. Re: Local Maximum and Minimum

Originally Posted by djnorris2000
Thanks, I can see there's a second solution, but I make it 2/3 pi while the book I'm working from makes it 5/6 pi and I'm not sure why I'm arriving at a different answer.
$\displaystyle \cos(2x) = \frac{1}{2} \Rightarrow 2x = \frac{\pi}{3}, ~ \frac{5 \pi}{3} \Rightarrow x = ....$