\(f(x) = log_4(x + \frac 1x)\)
Domain of f(x), \(x + \frac 1x > 0\)
⇒ \(\frac{x^2 + 1}x > 0\)
⇒ \(x \in (0, \infty)\) \((\because x^2 + 1 > 0)\)
\(\because x + \frac 1x = (\sqrt x)^2 + (\frac 1{\sqrt x})^2 - 2\sqrt x \times \frac 1{\sqrt x} + 2\)
\(= (\sqrt x - \frac1{\sqrt x})^2 + 2\)
\(\ge 2\)
\(\therefore log_4 (x + \frac 1x) \ge log_42\) (\(\because\) log is an increasing function)
⇒ \(f(x) \ge \frac 12 \) \(\left(\because log_42 = \frac{log\,2}{log\,4} = \frac{log\,2}{2log\,2} = \frac12\right)\)
⇒ \(f(x) \in [0.5, \infty)\).
