The greatest and least values of f(x)=tan ^{- | vedclass

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(B) Given $f(x) = 	an^{-1} x - \frac{1}{2} \ln x$ on the interval $x \in \left[\frac{1}{\sqrt{3}}, \sqrt{3}ight]$.First, find the derivative $f'(x)

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$f_{\max }=\pi / 6+\frac{1}{4} \ln 3$

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(B) Given $f(x) = 	an^{-1} x - \frac{1}{2} \ln x$ on the interval $x \in \left[\frac{1}{\sqrt{3}}, \sqrt{3}ight]$.First, find the derivative $f'(x)$:$f'(x) = \frac{1}{1+x^2} - \frac{1}{2x} = \frac{2x - (1+x^2)}{2x(1+x^2)} = \frac{-(x^2 - 2x + 1)}{2x(1+x^2)} = \frac{-(x-1)^2}{2x(1+x^2)}$.Since $-(x-1)^2 \leq 0$ and $2x(1+x^2) > 0$ for $x \in \left[\frac{1}{\sqrt{3}}, \sqrt{3}ight]$, we have $f'(x) \leq 0$ for all $x$ in the interval.Thus, $f(x)$ is a strictly decreasing function on the given interval.The maximum value occurs at the left endpoint $x = \frac{1}{\sqrt{3}}$:$f_{\max} = f\left(\frac{1}{\sqrt{3}}ight) = 	an^{-1}\left(\frac{1}{\sqrt{3}}ight) - \frac{1}{2} \ln\left(\frac{1}{\sqrt{3}}ight) = \frac{\pi}{6} - \frac{1}{2} \ln(3^{-1/2}) = \frac{\pi}{6} + \frac{1}{4} \ln 3$.The minimum value occurs at the right endpoint $x = \sqrt{3}$:$f_{\min} = f(\sqrt{3}) = 	an^{-1}(\sqrt{3}) - \frac{1}{2} \ln(\sqrt{3}) = \frac{\pi}{3} - \frac{1}{2} \ln(3^{1/2}) = \frac{\pi}{3} - \frac{1}{4} \ln 3$.

The greatest and least values of $f(x)=\tan ^{-1} x-\frac{1}{2} \ln x$ on $\left[\frac{1}{\sqrt{3}}, \sqrt{3}\right]$ are

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