The Corporeal Tutorial: Problem #9

This is quite a tricky problem. It incorporates the calculus approach to optimization but simultaneously plays upon a obscure method for evaluating trigonometric functions. But without further a due here is the question:

Find the absolute maximum and the absolute minimum of this function over the interval provided.

$f(x)=\frac{3-cosx}{sinx} ; \left [ \frac{\pi }{4},\frac{3\pi }{4} \right ]$

To start this question we quickly realize that we must differentiate and solve for zero. So lets see what we got:

$f(x)=\frac{3-cosx}{sinx} \therefore$

$f'(x)=\frac{(sinx\cdot sinx)-(3-cosx)(cosx)}{sin^{2}x}$

$f'(x)=\frac{sin^{2}x-3cosx+cos^{2}x}{sin^{2}x}$

$f'(x)=\frac{1-3cosx}{sin^{2}x}$

Alrighty then; now f(x) is undefined when sin^2=0 but all of these point are outside of our interval of consideration. Therefore we then find the points within the interval f(x)=0

$0=1-3cosx$

$x=cos^{-1}(\frac{1}{3})$

So the next thing we must do is evaluate the endpoints as well as our critical point.

$f(\frac{\pi }{4})=\frac{3-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=\frac{6}{\sqrt{2}}-1=3\sqrt{2}-1$

$f(cos^{-1}(\frac{1}{3}))=\frac{3-\frac{1}{3}}{\frac{2\sqrt{2}}{3}}=\frac{3}{\frac{2\sqrt{2}}{3}}-\frac{1}{2\sqrt{2}}=\frac{9}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}=2\sqrt{2}$

I skipped two parts to this:
1) in order to find the value of sinx at arcosx(1/3) one would have to draw up a triangle and use the Pythagorean theorem.
2) at the end of the calculation one must combined like denominators.

Onward!

$f(\frac{3\pi }{4})=\frac{3+\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=\frac{6}{\sqrt{2}}+1=3\sqrt{2}+1\therefore$

$f(\frac{3\pi }{4})> f(\frac{\pi }{4})> f(cos^{-1}\frac{1}{3})$

our conclusion would be that f(x) has an absolute maximum at $3\sqrt{2}+1$ and an absolute minimum at $2\sqrt{2}$

And thus answers our question being considered.

The Corporeal Tutorial

Tuesday, June 23, 2009

Problem #9

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