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Range

The minimum v...

Question

The minimum value of the function $6^{x} + 3^{x} + 6^{- x} + 3^{- x} + 2$ is

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Solution

Given $f (x) = 6^{x} + 3^{x} + 6^{- x} + 3^{- x} + 2$
$= (6^{x} + 6^{- x}) + (3^{x} + 3^{- x}) + 2 = (6^{x} + \frac{1}{6^{x}}) + (3^{x} + \frac{1}{3^{x}}) + 2$
using A.M-G.M inequality,
$(6^{x} + \frac{1}{6^{x}}) = (3^{x} + \frac{1}{3^{x}}) \geq 2$

$∴ f (x) \geq 6$

Range of the function is $[6, \infty)$
Thus the minimum value is $6$

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