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Question Number 80180 by mathocean1 last updated on 31/Jan/20
h(x)=((x−x^2 )/(x+1))  we defined this function on  R−{−1}→R    1) Study the variations of h then  draw up its table of variation.    please sirs i need your kind help
h(x)=xx2x+1wedefinedthisfunctiononR{1}R1)Studythevariationsofhthendrawupitstableofvariation.pleasesirsineedyourkindhelp
Answered by Rio Michael last updated on 31/Jan/20
 solution   −−−−−−−−−−  h(x) = ((x−x^2 )/(x−1))  h(x) = 0 ⇒ x −x^2  = 0 ⇒ x =0 or x = 1   h′(x) = (((x−x^2 )−(x−1)(1−2x))/((x−1)^2 )) = ((x −x^2  −(x−2x^2 −1+2x))/((x−1)^2 ))   h^′ (x) = ((x^2 −2x + 1)/((x−1)^2 ))    h^′ (x) > 0 ⇒ (x−1)^2  > 0  ⇒ x > 1   h is increasing for x > 1 and decreasing other wise
solutionh(x)=xx2x1h(x)=0xx2=0x=0orx=1h(x)=(xx2)(x1)(12x)(x1)2=xx2(x2x21+2x)(x1)2h(x)=x22x+1(x1)2h(x)>0(x1)2>0x>1hisincreasingforx>1anddecreasingotherwise
Commented by mathocean1 last updated on 31/Jan/20
Please sir h′ is not like this:  h′=(((1−2x)(x+1)−(x−x^2 ))/((x+1)^2 ))  ?
Pleasesirhisnotlikethis:h=(12x)(x+1)(xx2)(x+1)2?
Commented by mathocean1 last updated on 31/Jan/20
h(x)=((x−x^2 )/(x+1))
h(x)=xx2x+1
Commented by Rio Michael last updated on 31/Jan/20
sorry i copied the question wrongly
sorryicopiedthequestionwrongly
Commented by mathocean1 last updated on 31/Jan/20
please sir how can i draw up its   table of variation when   h′=((−x^2 −2x+1)/((x+1)^2 ))=−1 ?
pleasesirhowcanidrawupitstableofvariationwhenh=x22x+1(x+1)2=1?
Commented by JDamian last updated on 01/Feb/20
if  h(x) were ((x−x^2 )/(x−1)), then h(x)=((x(1−x))/(x−1))=−x  a very easy function
ifh(x)werexx2x1,thenh(x)=x(1x)x1=xaveryeasyfunction
Commented by mathocean1 last updated on 01/Feb/20
so... we can study h(x)=((x−x^2 )/(x+1))???
sowecanstudyh(x)=xx2x+1???

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