Mathematics - Application of Derivative Question with Solution | TestHub

$f(x)=\tan x-4x\Rightarrow f^{\prime }(x)={\sec }^{2}x-4$ When $\frac{-\pi }{3}<\mathrm{x}<\frac{\pi }{3},1<\sec \mathrm{x}<2$ Therefore, $1<{\sec }^{2}x<4$ $\Rightarrow -3<\left({\sec }^{2}x-4\right)<0$ Thus, for $\frac{-\pi }{3}<\mathrm{x}<\frac{\pi }{3},{\mathrm{f}}^{\prime }(\mathrm{x})<0$ Hence, $\mathrm{f}$ is strictly decreasing on $\left(\frac{-\pi }{3},\frac{\pi }{3}\right)$