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Question

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

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Solution

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

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