# Thread: Local Maximum and Minimum

1. ## Local Maximum and Minimum

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

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

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

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

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

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

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 $\pi$ radians apart on the Cosine graph, they will be $x=\frac{1}{2}\pi$ radians apart on the graph of Cos 2x, which would put the minimum point at $x=\frac{2}{3}\pi$ but the book I'm working from puts it at $x=\frac{5}{6}\pi$.

2. ## Re: Local Maximum and Minimum

You have determined the $x$-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.
$\cos(2x) = \frac{1}{2} \Rightarrow 2x = \frac{\pi}{3}, ~ \frac{5 \pi}{3} \Rightarrow x = ....$