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Question Number 76919 by jagoll last updated on 01/Jan/20

$$whatrange$$$$thefunctiony=\frac{x-1}{\sqrt{x^2+x}}?$$

Answered by MJS last updated on 01/Jan/20

$$\mathrm{defined}\mathrm{for}-\mathrm{\infty }<x<-1\vee 0<x<+\mathrm{\infty }$$$$\frac{d}{dx}\left[\frac{x-1}{\sqrt{x^2+x}}\right]=\frac{3x+1}{2{(x^2+x)}^{3/2}}=0\Rightarrow x=-\frac{1}{3}\Rightarrow$$$$\Rightarrow \mathrm{no}\mathrm{real}\mathrm{extremes}$$$$\underset{x\to -\mathrm{\infty }}{\lim }\frac{x-1}{\sqrt{x^2+x}}=-1$$$$\underset{x\to -1^{-}}{\lim }\frac{x-1}{\sqrt{x^2+x}}=-\mathrm{\infty }$$$$\underset{x\to 0^{+}}{\lim }\frac{x-1}{\sqrt{x^2+x}}=-\mathrm{\infty }$$$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{x-1}{\sqrt{x^2+x}}=1$$$$\Rightarrow \mathrm{range}\mathrm{is}-\mathrm{\infty }<y<1$$

Commented by jagoll last updated on 01/Jan/20

$$\mathrm{if}\mathrm{y}\mathrm{equal}\mathrm{to}1\Rightarrow \sqrt{{\mathrm{x}}^2+\mathrm{x}}=\mathrm{x}-1$$$$\mathrm{squaring}\mathrm{in}\mathrm{sides}\mathrm{both}$$$${\mathrm{x}}^2+\mathrm{x}={\mathrm{x}}^2-2\mathrm{x}+1\Rightarrow 3\mathrm{x}=1$$$$\mathrm{x}=\frac{1}{3}\mathrm{satisfy}\mathrm{in}\mathrm{domain}\mathrm{function}$$

Commented by john santu last updated on 01/Jan/20

$$sorrysir.squaringinsidesboth$$$$workifx-1⩾0.thevaluex-1<0forx=\frac{1}{3}$$

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