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Question Number 32489 by NECx last updated on 26/Mar/18
$${find}\:{the}\:{range}\:{of}\:{f}\left({x}\right)=\mathrm{1}+\sqrt{\mathrm{2}{x}−\mathrm{1}} \\ $$
Commented by prof Abdo imad last updated on 28/Mar/18
,+∞[ lim_(x→+∞) f(x)=+∞ and f([(1/2),+∞[ =[f((1/2)),^ +∞[ =[1 ,+∞[ .](Q32571.png)
$${D}_{{f}} =\left[\frac{\mathrm{1}}{\mathrm{2}},+\infty\left[\:\:{and}\:{f}^{'} \left({x}\right)\:=\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}{x}−\mathrm{1}}}\:\:>\mathrm{0}\:{on}\right]\frac{\mathrm{1}}{\mathrm{2}},+\infty\left[\right.\right. \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=+\infty\:\:{and}\: \\ $$$${f}\left(\left[\frac{\mathrm{1}}{\mathrm{2}},+\infty\left[\:=\left[{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\bar {,}\:+\infty\left[\:=\left[\mathrm{1}\:,+\infty\left[\:\:.\right.\right.\right.\right.\right.\right.\right. \\ $$
Answered by MJS last updated on 26/Mar/18
$$\mathrm{domain}:\:{x}\in\mathbb{R}\wedge{x}\geqq\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{range}:\:{x}\in\mathbb{R}\wedge{x}\geqq\mathrm{1} \\ $$