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Question Number 35769 by Tinkutara last updated on 23/May/18

Answered by ajfour last updated on 23/May/18

$$letf\left(x\right)=\frac{2x}{1+x^2}$$$$\frac{df\left(x\right)}{dx}=\frac{2(1+x^2)-4x^2}{{(1+x^2)}^2}=\frac{2(1-x)(1+x)}{{(1+x^2)}^2}$$$$\Rightarrow f{\left(x\right)}_{min}=f(-1)=-1$$$$f{\left(x\right)}_{max}=f\left(1\right)=1$$$$Hencey={\tan }^{-1}\left[f\left(x\right)\right]hasrange$$$$\in [-\frac{\pi }{4},\frac{\pi }{4}].$$

Commented by Tinkutara last updated on 23/May/18

Is f(x) increasing or decreasing continuously?

Commented by ajfour last updated on 23/May/18

decreasing, then increasing, then again decreasing.

Commented by Tinkutara last updated on 23/May/18

$$Thenwhyonlyf(-1)andf\left(1\right)?$$

Commented by ajfour last updated on 23/May/18

Commented by Tinkutara last updated on 23/May/18

Thank you very much Sir! I got the answer. ��������

Answered by tanmay.chaudhury50@gmail.com last updated on 24/May/18

$$y={tan}^{-1}\left(\frac{2x}{1+x^2}\right)$$$$tany=\frac{2x}{1+x^2}letx=tan\theta$$$$tany=\frac{2tan\theta }{1+{tan}^2\theta }$$$$tany=sin2\theta$$$$1⩾sin2\theta ⩾-1$$$$so1⩾tany⩾-1$$$$\frac{\mathrm{\Pi }}{4}⩾y⩾\frac{-\mathrm{\Pi }}{4}$$

Commented by Tinkutara last updated on 24/May/18

Thank you very much Sir! I got the answer. �������� Nice method.

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