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Question Number 32489 by NECx last updated on 26/Mar/18

$$findtherangeoff\left(x\right)=1+\sqrt{2x-1}$$

Commented by prof Abdo imad last updated on 28/Mar/18

$$D_f=[\frac{1}{2},+\mathrm{\infty }[and{f}^{{}^{\prime }}\left(x\right)=\frac{1}{\sqrt{2x-1}}>0on]\frac{1}{2},+\mathrm{\infty }[$$$${lim}_{x\to +\mathrm{\infty }}f\left(x\right)=+\mathrm{\infty }and$$$$f([\frac{1}{2},+\mathrm{\infty }[=[f\left(\frac{1}{2}\right)\overline{,}+\mathrm{\infty }[=[1,+\mathrm{\infty }[.$$

Answered by MJS last updated on 26/Mar/18

$$\mathrm{domain}:x\in \mathbb{R}\wedge x\geqq \frac{1}{2}$$$$\mathrm{range}:x\in \mathbb{R}\wedge x\geqq 1$$

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