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Question Number 73649 by aliesam last updated on 14/Nov/19
find the range    f(x)=(2/(6−(√(x+2))))
findtherangef(x)=26x+2
Commented by mathmax by abdo last updated on 14/Nov/19
x ∈D_f ⇔ x≥−2  and (√(x+2))≠6 ⇔x≥−2 and x+2 ≠36 ⇔x≥−2  and x≠34  ⇒D_f =[−2,34[∪]34,+∞[  f^′ (x)=2(−((−(1/(2(√(x+2)))))/((6−(√(x+2)))^2 ))) =(1/( (√(x+2))(6−(√(x+2)))^2 ))>0 ⇒f is increazing  on D_(f  )   we have f(−2)=(1/3) and  lim_(x→34^− ) f(x)   =+∞  lim_(x→34^+ )   f(x)=−∞ ⇒f([−2,34[)=[(1/3),+∞[ and  f(]34,+∞[) =]−∞,0[  variation of f  x                  −2                    34                           +∞  f^′                                  +         ∣∣                +  f                       (1/3)  inc +∞∣∣−∞      inc        0
xDfx2andx+26x2andx+236x2andx34Df=[2,34[]34,+[f(x)=2(12x+2(6x+2)2)=1x+2(6x+2)2>0fisincreazingonDfwehavef(2)=13andlimx34f(x)=+limx34+f(x)=f([2,34[)=[13,+[andf(]34,+[)=],0[variationoffx234+f+∣∣+f13inc+∣∣inc0
Commented by malwaan last updated on 14/Nov/19
x≥−2 and x≠ 34  ⇒−2≤ x < 34 ∨ x > 34  −2≤ x < 34⇒0≤x+2<36  ⇒0≤(√(x+2)) <6⇒0≥−(√(x+2)) >−6  ⇒6≥6−(√(x+2)) >0  ⇒(1/3)≤ (2/(6−(√(x+2)))) < ∞  ⇒(1/3) ≤ y < ∞ ....(1)  x > 34⇒34<x<∞  ⇒36< x+2 < ∞  ⇒6<(√(x+2)) <∞  ⇒−6>−(√(x+2)) >−∞  ⇒0>6−(√(x+2)) >−∞  ⇒−∞<(2/(6−(√(x+2)))) < 0  ⇒−∞<y<0 ....(2)  from(1);(2)  ⇒the range =  [1 ╱ 3 ; ∞[ ∪ ]−∞ ; 0 [
\boldsymbolx2\boldsymboland\boldsymbolx342\boldsymbolx<34\boldsymbolx>342\boldsymbolx<340\boldsymbolx+2<360\boldsymbolx+2<60\boldsymbolx+2>666\boldsymbolx+2>01326\boldsymbolx+2<13\boldsymboly<.(1)\boldsymbolx>3434<\boldsymbolx<36<\boldsymbolx+2<6<\boldsymbolx+2<6>\boldsymbolx+2>0>6\boldsymbolx+2><26\boldsymbolx+2<0<\boldsymboly<0.(2)\boldsymbolfrom(1);(2)\boldsymbolthe\boldsymbolrange=[13;[];0[
Answered by MJS last updated on 14/Nov/19
defined for x≥−2∧x≠34  f(−2)=(1/3)  lim_(x→34^− ) f(x)=+∞  lim_(x→34^+ ) f(x)=−∞  lim_(x→+∞) f(x)=0  ⇒ −∞<y≤0∨(1/3)≤y<+∞
definedforx2x34f(2)=13limx34f(x)=+limx34+f(x)=limx+f(x)=0<y013y<+
Commented by aliesam last updated on 14/Nov/19
god bless you sir
godblessyousir
Commented by aliesam last updated on 14/Nov/19
but i think that     −∞<y<0
butithinkthat<y<0
Commented by MJS last updated on 14/Nov/19
yes you are right, it′s a typo
yesyouareright,itsatypoyesyouareright,itsatypo

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