djeljenike

2021-08-30

Find the minimum and maximum values of $y=\sqrt{6}\theta -\sqrt{3}\mathrm{sec}\theta$
on the interval $\left[0,\frac{\text{pi}}{\text{3}}\right]$

Tuthornt

$f\left(0\right)=-\sqrt{6}$
$f\left(\frac{\pi }{3}\right)=\sqrt{6}\left(\frac{\pi }{3}\right)-\sqrt{3}\mathrm{sec}\left(\frac{\pi }{3}\right)=\sqrt{6}\left(\frac{\pi }{3}\right)-\sqrt{3}\left(2\right)=-0.899$
${f}^{\prime }\left(x\right)=\sqrt{6}-\sqrt{3}\mathrm{sec}\left(\theta \right)\mathrm{tan}\left(\theta \right)=0$
$\sqrt{6}=\sqrt{3}\mathrm{sec}\left(\theta \right)\mathrm{tan}\left(\theta \right)$
$\mathrm{sec}\left(\theta \right)\mathrm{tan}\left(\theta \right)=\frac{\sqrt{6}}{\sqrt{3}}$
$\theta =\frac{\pi }{4}$
$f\left(\frac{\pi }{4}\right)=\sqrt{6}\left(\frac{\pi }{4}\right)-\sqrt{3}\mathrm{sec}\left(\frac{\pi }{4}\right)=-0.52566$
we have:
$f\left(0\right)=-\sqrt{6}=-2.4495$
$f\left(\frac{\pi }{3}\right)=-0.899$
$f\left(\frac{\pi }{4}\right)=-0.52566$
minimum =
maximum =

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