The minimum value of the function f(x) = {x^{ | vedclass

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Question

$x^{10}+x^{2}+\frac{1}{x^{12}} \geq 3$ $(AM.GM)$

Equality occurs of $x=\pm 1$

But ${\sec ^{ - 1}}{\rm{x}}$ takes maximum at $\mathrm{x}=-$

$f

Vedclass · Accepted Answer

$\frac{{3\pi\  +\ 4}}{{\pi\  +\ 1}}$

Vedclass · Answer

$x^{10}+x^{2}+\frac{1}{x^{12}} \geq 3$ $(AM.GM)$

Equality occurs of $x=\pm 1$

But ${\sec ^{ - 1}}{\rm{x}}$ takes maximum at $\mathrm{x}=-$

$f_{\min }(\mathrm{x})=f(-1)=3+\frac{1}{\pi+1}$

The minimum value of the function $f(x) = {x^{10}} + {x^2} + \frac{1}{{{x^{12}}}} + \frac{1}{{\left( {1\ +\ {{\sec }^{ - 1}}\ x} \right)}}$ is

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