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let-f-x-2-x-1-2x-1-find-D-f-2-study-the-variation-of-f-x-3-calculate-1-3-f-x-dx-4-determine-f-1-x-and-calculate-1-3-f-1-x-dx-5-find-the-values-of-A-1-3-f

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Question Number 42631 by maxmathsup by imad last updated on 29/Aug/18
let f(x)=2(√(x−1)) −2x 1) find D_f 2) study the variation of f(x) 3 ) calculate ∫_1 ^3 f(x)dx 4) determine f^(−1) (x) and calculate ∫_1 ^3 f^(−1) (x)dx 5) find the values of A = ∫_1 ^3 ((f(x))/(f^(−1) (x)dx)) and B = ((∫_1 ^3 f(x))/(∫_1 ^3 f^(−1) (x))) dx.
letf(x)=2x−1−2x1)findDf2)studythevariationoff(x)3)calculate∫13f(x)dx4)determinef−1(x)andcalculate∫13f−1(x)dx5)findthevaluesofA=∫13f(x)f−1(x)dxandB=∫13f(x)∫13f−1(x)dx.
Commented by bbnnftm833 last updated on 30/Aug/18
we need the solution
weneedthesolution
Commented by maxmathsup by imad last updated on 31/Aug/18
1) x∈ D_f ⇔ x−1≥0 ⇒D_f =[1,+∞[ 2)lim_(x→+∞) f(x) =lim_(x→+∞) 2x( (√((x−1)/x^2 )) −1)=lim_(x→+∞) 2x((√((1/x)−(1/x^2 )))−1) =lim_(x→+∞) −2x =−∞ we have f^′ (x)= (1/( (√(x−1)))) −2 for x>1 ⇒f^′ (x)=((1−2(√(x−1)))/( (√(x−1)))) f^′ (x)=0 ⇔ 2(√(x−1))=1 ⇔(√(x−1))=(1/2) ⇔x−1=(1/4) ⇔x =(5/4) f^′ (x)>0 ⇔ 1−2(√(x−1))>0 ⇔2(√(x−1))<1 ⇔ 1<x<(5/4) f^′ (x)<0 ⇔ x>(5/4) x_ 1 (5/4) +∞ f^′ (x) + − f(x) −2 inc −(3/2) decr f(1)=−2 f((5/4))=2(√((5/4)−1))−2(5/4) =1−(5/2) =−(3/2)
1)x∈Df⇔x−1⩾0⇒Df=[1,+∞[2)limx→+∞f(x)=limx→+∞2x(x−1x2−1)=limx→+∞2x(1x−1x2−1)=limx→+∞−2x=−∞wehavef′(x)=1x−1−2forx>1⇒f′(x)=1−2x−1x−1f′(x)=0⇔2x−1=1⇔x−1=12⇔x−1=14⇔x=54f′(x)>0⇔1−2x−1>0⇔2x−1<1⇔1<x<54f′(x)<0⇔x>54x154+∞f′(x)+−f(x)−2inc−32decrf(1)=−2f(54)=254−1−254=1−52=−32
Commented by maxmathsup by imad last updated on 31/Aug/18
3) ∫_1 ^3 f(x)dx = ∫_1 ^3 {2(√(x−1))−2x}dx = 2 [(2/3)(x−1)^(3/2) ]_1 ^3 − [x^2 ]_1 ^3 =(4/3){ 2^(3/2) } −(9−1) = (4/3)(√2^3 ) −8 =(4/3)(2(√2)) −8 =((8(√2))/3) −8 .
3)∫13f(x)dx=∫13{2x−1−2x}dx=2[23(x−1)32]13−[x2]13=43{232}−(9−1)=4323−8=43(22)−8=823−8.
Commented by maxmathsup by imad last updated on 31/Aug/18
4) f(x)=y ⇔ x =f^(−1) (y) and f(x)=y ⇒2(√(x−1))−2x=y (x≥1) ⇒ 2(√(x−1)) =2x+y ⇒ 4(x−1) =(2x+y)^2 ⇒4x−4 =4x^2 +4xy +y^2 ⇒ 4x^2 +4xy +y^2 −4x+4 =0 ⇒ 4x^2 +4(y−1)x +y^2 +4 =0 Δ^′ =4(y−1)^2 −4(y^2 +4) =4y^2 −8y +4 −4y^2 −16 =−8y −12 this equation have solution ⇔−8y−12≥0 ⇔8y +12≤0 ⇔y ≤((−12)/8)(=−(3/2)) x_1 =((−2(y−1)+(√(−8y−12)))/4) =((−(y−1) +(√(−2y−3)))/2) x_2 =((−(y−1)−(√(−2y−3)))/2) x_1 −1 =((1−y +(√(−2y−3)))/2) −1 =((−1−y +(√(−2y −3)))/2) but y≤−(3/2) ⇒−y≥(3/2) ⇒−1−y ≥(1/2) ≥0 ⇒x_1 −1≥0 no need to verify x_2 ⇒f^(−1) (y) =((1−y +(√(−2y−3)))/2) ⇒f^(−1) (x) =((1−x+(√(−2x−3)))/2) with x≤−(3/2) .
4)f(x)=y⇔x=f−1(y)andf(x)=y⇒2x−1−2x=y(x⩾1)⇒2x−1=2x+y⇒4(x−1)=(2x+y)2⇒4x−4=4x2+4xy+y2⇒4x2+4xy+y2−4x+4=0⇒4x2+4(y−1)x+y2+4=0Δ′=4(y−1)2−4(y2+4)=4y2−8y+4−4y2−16=−8y−12thisequationhavesolution⇔−8y−12⩾0⇔8y+12⩽0⇔y⩽−128(=−32)x1=−2(y−1)+−8y−124=−(y−1)+−2y−32x2=−(y−1)−−2y−32x1−1=1−y+−2y−32−1=−1−y+−2y−32buty⩽−32⇒−y⩾32⇒−1−y⩾12⩾0⇒x1−1⩾0noneedtoverifyx2⇒f−1(y)=1−y+−2y−32⇒f−1(x)=1−x+−2x−32withx⩽−32.
Commented by maxmathsup by imad last updated on 31/Aug/18
4) forgive the Q is calculate ∫_(−3) ^(−2) f^(−1) (x)dx 5) the Q is find the value of A = ((∫_2 ^3 f(x)dx)/(∫_(−3) ^(−2) f^(−1) (x)dx)) B = ∫ ((f(x))/(f^(−1) (x)))dx .
4)forgivetheQiscalculate∫−3−2f−1(x)dx5)theQisfindthevalueofA=∫23f(x)dx∫−3−2f−1(x)dxB=∫f(x)f−1(x)dx.
Commented by maxmathsup by imad last updated on 31/Aug/18
4) we have f^(−1) (x)=((1−x+(√(−2x−3)))/2) ⇒ ∫_(−3) ^(−2) f^(−1) (x)dx =(1/2) ∫_(−3) ^(−2) (1−x) +(1/2) ∫_(−3) ^(−2) (√(−2x−3))dx but ∫_(−3) ^(−2) (1−x)dx =[x−(x^2 /2)]_(−3) ^(−2) =−2−2−(−3−(9/2))=−1 +(9/2) =(7/2) and ∫_(−3) ^(−2) (√(−2x−3))dx =[−(1/3)(−2x−3)^(3/2) ]_(−3) ^(−2) =−(1/3){1−(3)^(3/2) } =(√3) −(1/3) ⇒∫_(−3) ^(−2) f^(−1) (x)dx = (7/4) +((√3)/2) −(1/6) =((38)/(24)) +((√3)/2) =((19)/(12)) +((√3)/2)
4)wehavef−1(x)=1−x+−2x−32⇒∫−3−2f−1(x)dx=12∫−3−2(1−x)+12∫−3−2−2x−3dxbut∫−3−2(1−x)dx=[x−x22]−3−2=−2−2−(−3−92)=−1+92=72and∫−3−2−2x−3dx=[−13(−2x−3)32]−3−2=−13{1−(3)32}=3−13⇒∫−3−2f−1(x)dx=74+32−16=3824+32=1912+32
Commented by maxmathsup by imad last updated on 31/Aug/18
5) we have ∫_2 ^3 f(x)dx = 2 ∫_2 ^3 (√(x−1)) −∫_2 ^3 2x dx =2[(2/3)(x−1)^(3/2) ]_2 ^3 −[x^2 ]_2 ^3 =(4/3){ 2^(3/2) −1}−{9−4} =(4/3)(2(√2)) −5 =((8(√2))/3) −5 also we have ∫_(−3) ^(−2) f^(−1) (x)dx =((19)/(12)) +((√3)/2) ⇒ A = ((∫_2 ^3 f(x)dx)/(∫_(−3) ^(−2) f^(−1) (x)dx)) = ((((8(√2))/3)−5)/(((19)/(12)) +((√3)/2))) .
5)wehave∫23f(x)dx=2∫23x−1−∫232xdx=2[23(x−1)32]23−[x2]23=43{232−1}−{9−4}=43(22)−5=823−5alsowehave∫−3−2f−1(x)dx=1912+32⇒A=∫23f(x)dx∫−3−2f−1(x)dx=823−51912+32.
Answered by behi83417@gmail.com last updated on 30/Aug/18
1)x−1≥0⇒x≥1⇒D_f =[1,+∞) 2)y=2(√(x−1))−2x y′=(1/( (√(x−1))))−2⇒x≠1⇒D_f =(1,+∞) x−1=t^2 ⇒x=t^2 +1 f(t)=2t−2(t^2 +1)=−2(t^2 −t+1) ⇒t^2 −t+1=((−y)/2)⇒(t−(1/2))^2 =((−y)/2)−(3/4) ⇒t−(1/2)=±(√(−((2y+3)/4)))⇒t=(1/2)(1±(√(−2y−3))) ⇒(√(x−1))=(1/2)(1±(√(−2y−3))) ⇒x=1+(1/4)(1±(√(−2y−3)))^2 ⇒f^(−1) (x)=1+(1/4)(1±(√(−2x−3)))^2 −2y−3≥0⇒2y+3≤0⇒y≤−(3/2) ⇒R_f =(−∞,−(3/2)]. 3)∫_1 ^3 f(x)dx=∫_1 ^3 (2(√(x−1))−2x)dx= =[(4/3)(x−1)^(3/2) −x^2 ]_1 ^3 =8(((√2)/3)−1). 4)∫f^(−1) (x)dx=∫[1+(1/4)(1±(√(−2x−3)))^2 ]dx= =∫[1+(1/4)(1∓(2x+3)±2(√(−2x−3)))]dx I_1 =∫[1+(1/4)(2x+4+2(√(−2x−3)))]dx= =(1/2)(2x+(((x+1)^2 )/2)−(1/3)(−2x−3)^(3/2) )_1 ^3 = =(1/2)[6+8−27i−2−2+((i(√5))/3)]= =5−((81−(√5))/6).i
1)x−1⩾0⇒x⩾1⇒Df=[1,+∞)2)y=2x−1−2xy′=1x−1−2⇒x≠1⇒Df=(1,+∞)x−1=t2⇒x=t2+1f(t)=2t−2(t2+1)=−2(t2−t+1)⇒t2−t+1=−y2⇒(t−12)2=−y2−34⇒t−12=±−2y+34⇒t=12(1±−2y−3)⇒x−1=12(1±−2y−3)⇒x=1+14(1±−2y−3)2⇒f−1(x)=1+14(1±−2x−3)2−2y−3⩾0⇒2y+3⩽0⇒y⩽−32⇒Rf=(−∞,−32].3)∫31f(x)dx=∫31(2x−1−2x)dx=Missing \left or extra \right4)∫f−1(x)dx=∫[1+14(1±−2x−3)2]dx==∫[1+14(1∓(2x+3)±2−2x−3)]dxI1=∫[1+14(2x+4+2−2x−3)]dx=Missing \left or extra \right=12[6+8−27i−2−2+i53]==5−81−56.i

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