If the maximum value of $$a$$, for which the function $$f_a(x) = \tan^{-1}(2x) - 3ax + 7$$ is non-decreasing in $$\left(-\dfrac{\pi}{6}, \dfrac{\pi}{6}\right)$$, is $$\bar{a}$$, then $$f_{\bar{a}}\left(\dfrac{\pi}{8}\right)$$ is equal to

For

$$f_a(x)$$

to be non-decreasing on

$$\left(-\frac{\pi}{6},\frac{\pi}{6}\right)$$,

we must have

$$f_a'(x)\ge 0$$

throughout the interval.

Differentiating,

$$f_a'(x)=\frac{2}{1+4x^2}-3a$$

Therefore,

$$\frac{2}{1+4x^2}-3a\ge 0$$

$$3a\le \frac{2}{1+4x^2}$$

for every

$$x\in\left(-\frac{\pi}{6},\frac{\pi}{6}\right)$$.

To obtain the maximum possible value of

$$a$$,

we need the minimum value of

$$\frac{2}{1+4x^2}$$

on the interval.

Since

$$x^2$$

is largest at

$$x=\pm\frac{\pi}{6}$$,

we get

$$3\bar a=\frac{2}{1+\frac{\pi^2}{9}}$$

$$=\frac{18}{9+\pi^2}$$

Hence

$$\bar a=\frac{6}{9+\pi^2}$$

Now,

$$f_{\bar a}\left(\frac{\pi}{8}\right)=\tan^{-1}\left(\frac{\pi}{4}\right)-3\bar a\left(\frac{\pi}{8}\right)+7$$

Substituting

$$\bar a=\frac{6}{9+\pi^2}$$,

$$f_{\bar a}\left(\frac{\pi}{8}\right)=7+\tan^{-1}\left(\frac{\pi}{4}\right)-\frac{18}{9+\pi^2}\cdot\frac{\pi}{8}$$

$$=7+\tan^{-1}\left(\frac{\pi}{4}\right)-\frac{9\pi}{4(9+\pi^2)}$$

Therefore,

$$f_{\bar a}\left(\frac{\pi}{8}\right)=8-\frac{9\pi}{4(9+\pi^2)}$$