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Find the global minimum of the function $f(x) = \tan x + 3 \cot x$ on the interval $0 < x < \frac{\pi}{2}$. Write your answer as a multiple of $\pi$.

So when I graphed this equation I got a minimum of $x = 1.047$. How do I write this in terms of $\pi$, if I even did this right? Thanks!

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It is useful to put away the calculator. –  André Nicolas Feb 21 '13 at 20:49
    
Considering our teacher tells us to always graph the equation first I'd like to think the calculator is my best tool but thanks for the advice. So, if I did this right, the derivative would be sec^(x)+(3/tan(x)). I'm not really sure how to solve that. –  user56852 Feb 21 '13 at 20:52
    
Actually, the calculator points you to the correct answer. Doesn't that look like $\frac \pi 3$? –  Ross Millikan Feb 21 '13 at 20:55
    
wow, yes that makes sense! thanks! –  user56852 Feb 21 '13 at 20:57
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3 Answers 3

Substitue $y=\tan x$, then $y$ takes values in $(0,\infty)$ and $\cot x=\frac{1}{y}$.

Find the value of $y$ which makes $f=y+\frac{3}{y}$ minimal, and from there find $x$

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Since you’re finding an extremum, you (should) have $f\,'(x)=\sec^2x-3\csc^2x$. Setting this to $0$ and multiplying by $\sin^2x$ yields $\tan^2x-3=0$, or $\tan x=\pm\sqrt3$. (You’re permitted to multiply by $\sin^2x$ because it’s not $0$ at any point in the interval of interest.) Since $0<x<\frac{\pi}2$, you know that $\tan x>0$, so $\tan x=\sqrt 3$. This is a standard angle, one of the handful that you should simply know; what angle is it?

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Would that be 60 degrees or 240 degrees? Oh so that would make the minimum, 60 degrees, pi/3? –  user56852 Feb 21 '13 at 20:56
    
@user56852: Has to be $\pi/3$, not $4\pi/3$, since it’s between $0$ and $\pi/2$. Yes, $\pi/3$ is correct. –  Brian M. Scott Feb 21 '13 at 20:59
    
thank you a bunch! –  user56852 Feb 21 '13 at 20:59
    
@user56852: You’re welcome! –  Brian M. Scott Feb 21 '13 at 21:00
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This function tends to $+\infty$ at the bounds, and it is continuous, so it reaches a global minimum where its derivative is zero.

Now $$ f'(x)=\frac{1-4\cos^2x}{\sin^2x\cos^2x}. $$

Can you solve $f'(x)=0$ on $(0,\pi/2)$?

Since there is a unique solution, this must be your minimum.

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