MINIMUM! :D

Algebra Level 3

find the minimum value of
${ \tan }^{ 2 }x + 2 \cot x$ when $x$ is in the interval $(0,\frac { \pi }{ 2 } )$

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0.5 2.5 1 3

1 solution

Brian Charlesworth
Dec 23, 2014

Since both $\tan(x)$ and $\cot(x)$ are greater than zero on the given interval, we can use the AM-GM inequality to find that

$\dfrac{\tan^{2}(x) + \cot(x) + \cot(x)}{3} \ge \sqrt[3]{\tan^{2}(x)*\cot(x)*\cot(x)} = 1$

$\Longrightarrow (\tan^{2}(x) + 2\cot(x)) \ge 3$ .

The minimum of $\boxed{3}$ is achieved when $x = \frac{\pi}{4}$ .

What is AM-GM inequality?

Sridhar Muruganandham - 6 years, 5 months ago

You can read about this inequality in this wiki .

Brian Charlesworth - 6 years, 5 months ago

What if I put the value as : x=30° . Then what?

Yashasvi Goel - 6 years, 5 months ago

Plugging in $x = 30^{\circ}$ gives a value of $3.797....$ for the expression, which exceeds the value of $3$ achieved at $x = 45^{\circ}$ , and hence is not the minimum.

Brian Charlesworth - 6 years, 5 months ago

MINIMUM! :D

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