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The Problem

Find the absolute minimum and maximum values of the following function on the given interval: $\ f(\theta) = \cos \theta; -\pi \leq \theta \leq {\pi \over 6} $

What I've Done So Far

  1. Find the derivative of $\ f(\theta) $. This equals $\ -\sin \theta $.
  2. Determine critical points. It won't ever be undefined, but I know there are points within the interior of the domain where the function is equal to zero.

So where I'm stuck is finding at what points $\ f(\theta) $ equals zero. Is it as simple as taking the inverse $\ \sin $ of $\ \theta $?

Thanks! Any help will be much appreciated!

Garren

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  • $\begingroup$ You could just check it from the graph of the cosine function $\endgroup$ – Luca Oct 28 '16 at 19:37
  • $\begingroup$ True. But I would like to learn from the problem so that I can also find the answer algebraically. $\endgroup$ – Garren Miller Oct 28 '16 at 19:38
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If we have a function $f$ then at the maxima and minima of $f$, we have $f'=0$. $$f(\theta)=\cos(x)\implies f'(\theta)=-\sin(\theta)$$ We know that: $$-\sin(\theta)=0\implies \sin(\theta)=0\implies \theta=\pi k,k\in\Bbb{Z}$$ Since we have a domain $-\pi\leq\theta\leq\dfrac{\pi}{6}$, we have maxima or minima at $0\pi=0$ and $-1\pi=-\pi$. And: $$\cos(0)=1$$ $$\cos(-\pi)=-1$$ So we have a maximum at $(0,1)$ and a minimum at $(-\pi,-1)$. Since there are no other multiples of $\pi$ in the interval $-\pi\leq\theta\leq\dfrac{\pi}{6}$ and therefore no other solutions to $-\sin(\theta)=\theta$ in the interval, we are done.

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