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Question Number 133989 by Mamifere last updated on 26/Feb/21

$$\frac{x}{x2+1}<arctan\left(x\right)<x$$ $$$$

Answered by Ñï= last updated on 26/Feb/21

$$Letf\left(x\right)={tan}^{-1}x$$ $$wehavef\left(x\right)-f\left(0\right)=f{\left(\xi \right)}^{\prime }(x-0)$$ $${tan}^{-1}x=f{\left(\xi \right)}^{\prime }x$$ $$f{\left(\xi \right)}^{\prime }=\frac{1}{1+{\xi }^2}$$ $$1)::Ifx>0Then0<\xi <x$$ $$\therefore 1<1+{\xi }^2<1+x^2$$ $$\Rightarrow \frac{1}{1+x^2}<\frac{1}{1+{\xi }^2}<1$$ $$\Rightarrow \frac{x}{1+x^2}<\frac{x}{1+{\xi }^2}<x$$ $$\therefore \frac{x}{1+x^2}<{tan}^{-1}x<x$$ $$2)::Ifx<0Thenx<\xi <0$$ $$\therefore 1<1+{\xi }^2<1+x^2$$ $$\Rightarrow \frac{1}{1+x^2}<\frac{1}{1+{\xi }^2}<1$$ $$\Rightarrow \frac{x}{1+x^2}>\frac{x}{1+{\xi }^2}>x$$ $$\therefore \frac{x}{1+x^2}>{tan}^{-1}x>x$$

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