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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)))) ?
whatrangethefunctiony=x1x2+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
definedfor<x<10<x<+ddx[x1x2+x]=3x+12(x2+x)3/2=0x=13norealextremeslimxx1x2+x=1limx1x1x2+x=limx0+x1x2+x=limx+x1x2+x=1rangeis<y<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
ifyequalto1x2+x=x1squaringinsidesbothx2+x=x22x+13x=1x=13satisfyindomainfunction
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)
sorrysir.squaringinsidesbothworkifx10.thevaluex1<0forx=13

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