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Question Number 59788 by aliesam last updated on 14/May/19
find the local minimum and maximum value a^2 y=x^2 (a−x) f(x)=(4/(2−x))+(9/(x−3))
findthelocalminimumandmaximumvaluea2y=x2(a−x)f(x)=42−x+9x−3
Commented by kaivan.ahmadi last updated on 14/May/19
y=((x^2 (a−x))/a^2 ) and let a>0 y′=((2ax−3x^2 )/a^2 )=0⇒x(2a−3x)=0⇒ { ((x=0)),((x=((2a)/3))) :} { ((x<0⇒f′(x)<0)),((0<x<((2a)/3)⇒f′(x)>0)),((x>((2a)/3)⇒f′(x)<0)) :} ⇒x=0 is local min and x=((2a)/3) is local max if a<0 then x=0 is local max and x=((2a)/3) is local min
y=x2(a−x)a2andleta>0y′=2ax−3x2a2=0⇒x(2a−3x)=0⇒{x=0x=2a3{x<0⇒f′(x)<00<x<2a3⇒f′(x)>0x>2a3⇒f′(x)<0⇒x=0islocalminandx=2a3islocalmaxifa<0thenx=0islocalmaxandx=2a3islocalmin
Commented by aliesam last updated on 14/May/19
thank you sir i am waiting number to solution
thankyousiriamwaitingnumbertosolution
Commented by kaivan.ahmadi last updated on 14/May/19
f(x)=(4/(2−x))+(9/(x−3))⇒ D_f =R−{2,3} f′(x)=(4/((2−x)^2 ))−(9/((x−3)^2 ))=0⇒ 4(x−3)^2 =9(2−x)^2 ⇒ 4x^2 −24x+36=9x^2 −36x+36⇒ 5x^2 −12x=0⇒x(5x−12)=0⇒ { ((x=0)),((x=((12)/5))) :} x<0⇒f′(x)<0 0<x<((12)/5)⇒f′(x)>0 x>((12)/5)⇒f′(x)<0 ⇒x=0 min and x=((12)/5)max
f(x)=42−x+9x−3⇒Df=R−{2,3}f′(x)=4(2−x)2−9(x−3)2=0⇒4(x−3)2=9(2−x)2⇒4x2−24x+36=9x2−36x+36⇒5x2−12x=0⇒x(5x−12)=0⇒{x=0x=125x<0⇒f′(x)<00<x<125⇒f′(x)>0x>125⇒f′(x)<0⇒x=0minandx=125max
Commented by aliesam last updated on 14/May/19
thank you sir
thankyousir
Answered by meme last updated on 14/May/19
y=((x^2 (a−x))/a^2 ). a≠0 y^′ =((2x(a−x)−x^2 )/a^2 )=0 2x(a−x)−x^2 =0 2(a−x)=x 2a=3x x=((2a)/3) if x∈]−∞;((2a)/3)]y^′ ≤0.y has minimal if x∈[((2a)/3);+∞[y^′ ≥0.y has maximal
y=x2(a−x)a2.a≠0y′=2x(a−x)−x2a2=02x(a−x)−x2=02(a−x)=x2a=3xx=2a3ifx∈]−∞;2a3]y′⩽0.yhasminimalifx∈[2a3;+∞[y′⩾0.yhasmaximal
Answered by MJS last updated on 14/May/19
y=−(1/a^2 )x^3 +(1/a)x^2 y′=−(3/a^2 )x^2 +(2/a)x=0 ⇒ x=0∨x=((2a)/3) y′′=−(6/a^2 )x+(2/a) x=0 ⇒ y′′=(2/a) ⇒ max with a<0; min with a>0 x=((2a)/3) ⇒ y′′=−(2/a) ⇒ min with a<0; max with a>0 f(x)=(4/(2−x))+(9/(x−3))=((5x−6)/((x−3)(x−2))) f′(x)=−((x(5x−12))/((x−3)^2 (x−2)^2 ))=0 ⇒ x=0∨x=((12)/5) f′′(x)=((2(5x^3 −18x^2 +36))/((x−3)^3 (x−2)^3 )) f′′(0)=(1/3)>0 ⇒ min at x=0 f′′(((12)/5))=−((625)/3) ⇒ max at x=((12)/5)
y=−1a2x3+1ax2y′=−3a2x2+2ax=0⇒x=0∨x=2a3y″=−6a2x+2ax=0⇒y″=2a⇒maxwitha<0;minwitha>0x=2a3⇒y″=−2a⇒minwitha<0;maxwitha>0f(x)=42−x+9x−3=5x−6(x−3)(x−2)f′(x)=−x(5x−12)(x−3)2(x−2)2=0⇒x=0∨x=125f″(x)=2(5x3−18x2+36)(x−3)3(x−2)3f″(0)=13>0⇒minatx=0f″(125)=−6253⇒maxatx=125

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