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

what range the function y = ((x−1)/(√(x^2 +x))) ?

$${what}\:{range}\: \\ $$$${the}\:{function}\:{y}\:=\:\frac{{x}−\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +{x}}}\:? \\ $$

Answered by MJS last updated on 01/Jan/20

defined for −∞<x<−1∨0<x<+∞ (d/dx)[((x−1)/(√(x^2 +x)))]=((3x+1)/(2(x^2 +x)^(3/2) ))=0 ⇒ x=−(1/3) ⇒ ⇒ no real extremes lim_(x→−∞) ((x−1)/(√(x^2 +x)))=−1 lim_(x→−1^− ) ((x−1)/(√(x^2 +x)))=−∞ lim_(x→0^+ ) ((x−1)/(√(x^2 +x)))=−∞ lim_(x→+∞) ((x−1)/(√(x^2 +x)))=1 ⇒ range is −∞<y<1

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

Commented by jagoll last updated on 01/Jan/20

if y equal to 1 ⇒ (√(x^2 +x )) = x−1 squaring in sides both x^2 +x = x^2 −2x+1 ⇒ 3x = 1 x = (1/3) satisfy in domain function

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

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

sorry sir. squaring in sides both work if x−1 ≥0. the value x−1 < 0 for x = (1/3)

$${sorry}\:{sir}.\:{squaring}\:{in}\:{sides}\:{both}\: \\ $$$${work}\:{if}\:{x}−\mathrm{1}\:\geqslant\mathrm{0}.\:\:{the}\:{value}\:{x}−\mathrm{1}\:<\:\mathrm{0}\:\:{for}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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