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Question Number 32776 by NECx last updated on 02/Apr/18
Find the optimum points of  the function y=f(x)    f(x)=2x^3 −3x^2 −36x+34
Findtheoptimumpointsofthefunctiony=f(x)f(x)=2x33x236x+34
Answered by MJS last updated on 02/Apr/18
f′(x)=0∧f′′(x)>0⇒min  f′(x)=0∧f′′(x)<0⇒max    f′(x)=6x^2 −6x−36  f′′(x)=12x−6    f′(x)=0  x^2 −x−6=0  x=(1/2)±(√((1/4)+6))=(1/2)±(√((25)/4))=  =(1/2)±(5/2)⇒x_1 =−2; x_2 =3  f′′(−2)=−30⇒max= (((−2)),((78)) )  f′′(3)=30⇒min= ((3),((−47)) )
f(x)=0f(x)>0minf(x)=0f(x)<0maxf(x)=6x26x36f(x)=12x6f(x)=0x2x6=0x=12±14+6=12±254==12±52x1=2;x2=3f(2)=30max=(278)f(3)=30min=(347)
Commented by NECx last updated on 02/Apr/18
thank you so much
thankyousomuch

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