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Question Number 41136 by math khazana by abdo last updated on 02/Aug/18
let f(x)=2x−(√(x−1)) find ∫ f^(−1) (x)dx .
letf(x)=2x−x−1find∫f−1(x)dx.
Answered by MJS last updated on 02/Aug/18
f(x) defined for x∈[1; +∞] f′(x)=2−(1/(2(√(x−1))))=0 ⇒ x=((17)/(16)) f′′(x)=(1/(4(x−1)^(3/2) )); f′′(((17)/(16)))=16>0 ⇒ min at ((((17)/(16))),(((15)/8)) ) range(f(x))=[((15)/8); +∞[ x=2y−(√(y−1)) (√(y−1))=2y−x y−1=4y^2 −4xy+x^2 y^2 −(x+(1/4))y+((x^2 +1)/4)=0 y_1 =(x/2)+(1/8)−((√(8x−15))/8) defined for x∈[((15)/8); +∞] y_1 ′=(1/2)−(1/(2(√(8x−15))))=0 ⇒ x=2 y_1 ′′=(2/((8x−15)^(3/2) )); x=2 ⇒ y_2 ′′=2>0 ⇒ min at ((2),(1) ) range=[1; +∞] y_2 =(x/2)+(1/8)+((√(8x−15))/8) defined for x∈[((15)/8); +∞] y_2 ′=(1/2)+(1/(2(√(8x−15))))=0 ⇒ no solution range=[((17)/(16)); +∞] f^(−1) (x)= { (((x/2)+(1/8)−((√(8x−15))/8); x∈[((15)/8); 2])),(((x/2)+(1/8)+((√(8x−15))/8); x∈]((15)/8); +∞[)) :} ∫f^(−1) (x)dx= { (((x^2 /4)+(x/8)−(((8x−15)^(3/2) )/(96)); x∈[((15)/8); 2])),(((x^2 /4)+(x/8)+(((8x−15)^(3/2) )/(96)); x∈]((15)/8); +∞[)) :} but of course if you want the area between this function and the x−axis you must take ∫f^(−1) (x)dx= { (((x^2 /4)+(x/8)−(((8x−15)^(3/2) )/(96)); x∈[((15)/8); 2])),(((x^2 /4)+(x/8)+(((8x−15)^(3/2) )/(96)); x∈]2; +∞[)) :}
f(x)definedforx∈[1;+∞]f′(x)=2−12x−1=0⇒x=1716f″(x)=14(x−1)32;f″(1716)=16>0⇒minat(1716158)range(f(x))=[158;+∞[x=2y−y−1y−1=2y−xy−1=4y2−4xy+x2y2−(x+14)y+x2+14=0y1=x2+18−8x−158definedforx∈[158;+∞]y1′=12−128x−15=0⇒x=2y1″=2(8x−15)32;x=2⇒y2″=2>0⇒minat(21)range=[1;+∞]y2=x2+18+8x−158definedforx∈[158;+∞]y2′=12+128x−15=0⇒nosolutionrange=[1716;+∞]f−1(x)={x2+18−8x−158;x∈[158;2]x2+18+8x−158;x∈]158;+∞[∫f−1(x)dx={x24+x8−(8x−15)3296;x∈[158;2]x24+x8+(8x−15)3296;x∈]158;+∞[butofcourseifyouwanttheareabetweenthisfunctionandthex−axisyoumusttake∫f−1(x)dx={x24+x8−(8x−15)3296;x∈[158;2]x24+x8+(8x−15)3296;x∈]2;+∞[
Commented by math khazana by abdo last updated on 03/Aug/18
thank you sir mjs for this hard work .
thankyousirmjsforthishardwork.

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