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Let $f(\theta) =\cos^2(\theta)-2\sin(\theta)$, find the local maximums, local minimums, or neither.

My solution is:

$y'=-2\sin(\theta)\cos(\theta)-2\cos(\theta)=0\Rightarrow \theta=2k\pi+\pi /2, 2k\pi+3\pi/2$.

$y''=-2\cos(2\theta)+2\sin(\theta)\Rightarrow \theta=2k\pi+\pi/2,y''>0;\theta=2k\pi+3\pi/2,y''=0$

Thus, the result is min at $x=2k\pi+\pi/2$.

But the Right Answer also includes: max at $x=2k\pi+3\pi/2$. What is wrong with my solution? Thanks!!!

zhaoyin.usm
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  • Instead of writing the trigonometric functions as $cos$ or $sin$ - by typing: cos or sin -, write them as $\cos$ or $\sin$ - by typing \cos or \sin. –  Dec 30 '13 at 11:42
  • If $y'' = 0$, then the second derivative test is inconclusive. Perhaps you can use the first derivative test at $\theta = 2k\pi + 3\pi/2$. – Prahlad Vaidyanathan Dec 30 '13 at 11:43
  • @ Wanderer: you have to discuss the sign of the first derivative – Matheman Dec 30 '13 at 11:56
  • Great! I got it. It should be back to analysis the first derivative. Thank you! – zhaoyin.usm Dec 30 '13 at 11:56

1 Answers1

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For the maxima, show that $$f'''(2k\pi + \frac{3\pi}2) = 0$$ as well and see $$f^{(4)}(2k\pi + \frac{3\pi}2) < 0$$

In general, if you have all derivatives up to $f'(x^\ast) = \ldots = f^{(k-1)}(x^\ast) = 0$ and $f^{(k)}(x^\ast) \neq 0$ then

  • If $k$ is even and $f^{(k)}(x^\ast) > 0$, it's a minimum
  • If $k$ is even and $f^{(k)}(x^\ast) < 0$, it's a maximum
  • If $k$ is odd, it's a saddle point
AlexR
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