If $a < 0$,then the function $f(x) = e^{ax} + e^{-ax}$ is monotonically decreasing for all values of $x$ where ...

Let $I$ be any interval such that $I \cap [-1, 1] = \phi$. Prove that the function $f$ given by $f(x) = x + \frac{1}{x}$ is strictly increasing on $I$.

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

If $f(x) = kx^3 - 9x^2 + 9x + 3$ is monotonically increasing in each interval,then

For every value of $x$,the function $f(x)=\frac{1}{a^{x}}, a>0$ is

If $f(x) = \frac{x}{\log x}$,then $f(x)$ is increasing in