I'm self teaching and doing a book exercise which asks: "Considering only positive values of x, locate the first two turning points on the curve $2\sin x - x$ and determine whether they are maximum or minimum points"

My first step was to set the derivative to $0$ and solve:

$\frac{d}{dx}(2\sin x - x) = 2\cos x - 1$.

$2\cos x - 1 = 0 \Rightarrow \cos x = \frac{1}{2} \Rightarrow x = \frac{\pi}{3}$

To determine max/min I plugged $\frac{\pi}{3}$ into the second derivative:

$\frac{d}{dx}(2\cos x - 1) = -2\sin x$

$-2\sin \frac{\pi}{3} = -\sqrt{3}$ which is negative, so $\frac{\pi}{3}$ is a maximum point.

I then sketched $2\cos x - 1$ and figured the next turning point (i.e. when derivative crosses the $x$-axis) would be $2\pi - \frac{\pi}{3} = \frac{5}{3}\pi$ and figured as the previous point was a max, then this would be a min. (I felt like I was guessing a bit here and would have preferred to discover that analytically). My answers are therefore:

$\dfrac{\pi}{3}$ (max), $\dfrac{5\pi}{3}$ (min)

The back of the book gives these answers:

($\frac{\pi}{3}$, $\sqrt{3} - \frac{1}{3}\pi$), max

($\frac{5\pi}{3}$, $-\sqrt{3} - \frac{5}{3}\pi$), min

My question is where do these solutions that involve $\sqrt{3}$ come from? I did consider it to be a book error as it asks for first two turning points but there's 4 answers there.

  • $\begingroup$ I think they are (confusingly) just including the function value as well. $\endgroup$ – copper.hat Apr 25 '13 at 16:07

Your answer is right.

The answers given are the points on the Cartesian plane $(x,f(x))$ whereas you have found only the abscissa.Here your $f(x)=2\sin x -x$

  • $\begingroup$ DOH! I feel very silly now. Many thanks. $\endgroup$ – PeteUK Apr 25 '13 at 16:10
  • $\begingroup$ Nevermind. It's all right! Got a good typing practice though. :D $\endgroup$ – ABC Apr 25 '13 at 16:11

A turning point is an $(x,y)$ pair. You found the $x$-values, correctly. To find the $y$-values, plug into $2\sin x -x$.


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