obrozenecy6

Answered question

2022-01-15

Find the minimum value of f

My approach is as follows. I tried to solve it by segregating it
$f\left(x\right)=\frac{1}{\sqrt{3}\mathrm{tan}x}+\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)\frac{1}{\sqrt{3}-\mathrm{tan}x}$
but f′(x) is getting more and more complicated

Answer & Explanation

Joseph Lewis

Beginner2022-01-16Added 43 answers

$g\left(x\right)=1+\frac{1}{2\mathrm{sin}x\mathrm{cos}\left(x+\frac{\pi }{6}\right)}=1+\frac{1}{\mathrm{sin}\left(2x+\frac{\pi }{6}\right)-\mathrm{sin}\frac{\pi }{6}}$
Now we need to maximize
$\mathrm{sin}\left(2x+\frac{\pi }{6}\right)$
in
$\left(\frac{\pi }{6},\frac{2\pi }{3}+\frac{\pi }{6}\right)$

Mason Hall

Beginner2022-01-17Added 36 answers

The first derivate of your function is:
${f}^{\prime }\left(x\right)=\mathrm{cot}\left(x\right){\mathrm{sec}\left(x+\frac{\pi }{6}\right)}^{2}-\mathrm{tan}\left(x+\frac{\pi }{6}\right){\mathrm{csc}\left(x\right)}^{2}$
Now, we have to impose f′(x)=0 and so:
$\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)-\mathrm{sin}\left(x+\frac{\pi }{6}\right)\mathrm{cos}\left(x+\frac{\pi }{6}\right)=0$
That can be rewritten as:
$\mathrm{sin}\left(2x\right)=\mathrm{sin}\left(2x+\frac{\pi }{3}\right)$
We know that:
$\mathrm{sin}\left(\alpha \right)=\mathrm{sin}\left(\beta \right)↔\alpha +\beta =\pi +2k\pi \vee \alpha =\beta +2k\pi$
So we have:

The only solution is $x=\frac{\pi }{6}$. So:
$f\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}=3$

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