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Question Number 130354 by greg_ed last updated on 24/Jan/21

$$\mathrm{f}\left(x\right)=\frac{4x^2+1}{2x^2+1}$$$$\mathrm{prove}\mathrm{that}1⩽\mathrm{f}\left(x\right)⩽2$$

Answered by MJS_new last updated on 24/Jan/21

$$f\left(x\right)=2-\frac{1}{2x^2+1}$$$$\mathrm{minimum}\mathrm{of}\frac{1}{2x^2+1}\mathrm{is}0(\mathrm{with}x=\pm \mathrm{\infty })$$$$\mathrm{maximum}\mathrm{of}\frac{1}{2x^2+1}\mathrm{is}1(\mathrm{with}x=0)$$$$\Rightarrow$$$$\mathrm{minimum}\mathrm{of}f\left(x\right)\mathrm{is}2-1=1$$$$\mathrm{maximum}\mathrm{if}f\left(x\right)\mathrm{is}2-0=2$$

Answered by ajfour last updated on 24/Jan/21

$$f\left(x\right)=\frac{4(x^2+\frac{1}{2})-1}{2(x^2+\frac{1}{2})}$$$$=2-\frac{\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)+x^2}$$$$\Rightarrow 1⩽f\left(x\right)⩽2$$$$$$

Answered by nueron last updated on 24/Jan/21

Answered by Olaf last updated on 24/Jan/21

$$f\left(x\right)=\frac{4x^2+1}{2x^2+1}=2-\frac{1}{2x^2+1}$$$$\Rightarrow f\left(x\right)⩽2\left(\mathrm{trivial}\right)$$$$f\left(x\right)-1=\frac{4x^2+1}{2x^2+1}-1=\frac{2x^2}{2x^2+1}⩾0$$$$\Rightarrow f\left(x\right)⩾1\left(\mathrm{trivial}\right)$$

Answered by john_santu last updated on 25/Jan/21

$$letf\left(x\right)=y$$$$\Rightarrow 2x^{2}y-4x^2+y-1=0$$$$\Rightarrow (2y-4)x^2+y-1=0$$$$D=0-4.(y-1)(2y-4)⩾0$$$$\Rightarrow (y-1)(2y-4)⩽0$$$$\Rightarrow 1⩽y⩽2\Rightarrow 1⩽f\left(x\right)⩽2$$

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