Question

Show that the function f(x) = tan x − 4x is decreasing function on (−π/3, π/3).

Answer 1

$$\begin{gathered} f\left(x\right)=\tan x-4x \\ f\text{'}\left(x\right)={\sec }^{2}x-4 \\ =\frac{1}{{\cos }^{2}x}-4 \\ =\frac{1-4{\cos }^{2}x}{{\cos }^{2}x} \\ ={\sec }^{2}x\left(1-4{\cos }^{2}x\right) \\ =4{\sec }^{2}x\left(\frac{1}{4}-{\cos }^{2}x\right) \\ =4{\sec }^{2}x\left(\frac{1}{2}-\cos x\right)\left(\frac{1}{2}+\cos x\right) \\ \text{Here,} \\ \frac{-\mathrm{\pi }}{3}<x<\frac{\mathrm{\pi }}{3} \\ \cos x>\frac{1}{2} \\ \Rightarrow -\cos x<\frac{-1}{2} \\ \Rightarrow \frac{1}{2}-\cos x<0...\left(1\right) \\ \mathrm{Also}, \\ \frac{-\mathrm{\pi }}{3}<x<\frac{\mathrm{\pi }}{3}\Rightarrow \cos x>\frac{1}{2} \\ \Rightarrow \frac{1}{2}+\cos x>1 \\ \Rightarrow \frac{1}{2}+\cos x>0...\left(2\right) \\ 4{\sec }^{2}x>0\left[\because a^2>0\right] \\ \mathrm{Now}, \\ f\text{'}\left(x\right)=4{\sec }^{2}x\left(\frac{1}{2}-\cos x\right)\left(\frac{1}{2}+\cos x\right)<0,\forall \in \left(\frac{-\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{3}\right)\left[\text{From eqs. (1) and (2)}\right] \\ \text{So,}f\left(x\right)\text{is decreasing on}\left(\frac{-\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{3}\right)\text{.} \end{gathered}$$