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Let-y-f-x-f-0-5-f-x-3x-2-0-2-f-x-dx-Find-the-maximum-value-of-y-f-x-

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Question Number 25397 by naka3546 last updated on 09/Dec/17
Let y = f (x) f (0) = 5 f ′(x) = −3x + 2 ∫_0 ^2 f (x) dx Find the maximum value of y = f(x) .
Lety=f(x)f(0)=5f′(x)=−3x+2∫02f(x)dxFindthemaximumvalueofy=f(x).
Commented by prakash jain last updated on 09/Dec/17
c=2∫_0 ^2 f(x)dx f^′ (x)=−3x+c f(x)=−((3x^2 )/2)+cx+c_1 f(0)=5⇒c_1 =5 f(x)=−((3x^2 )/2)+cx+5 2∫_0 ^( 2) f(x)dx=2[−(x^3 /2)+((cx^2 )/2)+5x]_0 ^2 c=−8+4c+20 c=−4 f(x)=−((3x^2 )/2)−4x+5 quadratic with a<0 so maximum value at x=((−b)/(2a)) x=(4/(−3))=−(4/3) f_(max) (x)=−(3/2)((4/3))^2 +((16)/3)+5 =−(8/3)+((16)/3)+5=((23)/3)
c=2∫02f(x)dxf′(x)=−3x+cf(x)=−3x22+cx+c1f(0)=5⇒c1=5f(x)=−3x22+cx+52∫02f(x)dx=2[−x32+cx22+5x]02c=−8+4c+20c=−4f(x)=−3x22−4x+5quadraticwitha<0somaximumvalueatx=−b2ax=4−3=−43fmax(x)=−32(43)2+163+5=−83+163+5=233
Commented by abwayh last updated on 09/Dec/17
another way; put f(x)=Ax^2 +Bx+c f(0)=c=5 f ^� (x)=2Ax+B ∵ f ^� (x)=−3x+2∫_0 ^2 f(x)dx ∴2Ax+B=−3x+2∫_0 ^2 (Ax^2 +Bx+c)dx 2Ax+B=−3x+2[((Ax^3 )/3) +((Bx^2 )/2) +cx]_0 ^2 2Ax+B=−3x+((16A)/3) +4B +4c (by compairing) 2A=−3 ⇒ A=((−3)/2) B=((16A)/3) +4B+4c B=((16)/3)(((−3)/2))+4B+4(5) B=−8+4B+20 ⇒3B=−12 ⇒B=−4 ∴f(x)=((−3)/2) x^2 −4x+5 continue
anotherway;putf(x)=Ax2+Bx+cf(0)=c=5f´(x)=2Ax+B∵f´(x)=−3x+2∫02f(x)dx∴2Ax+B=−3x+2∫02(Ax2+Bx+c)dx2Ax+B=−3x+2[Ax33+Bx22+cx]022Ax+B=−3x+16A3+4B+4c(bycompairing)2A=−3⇒A=−32B=16A3+4B+4cB=163(−32)+4B+4(5)B=−8+4B+20⇒3B=−12⇒B=−4∴f(x)=−32x2−4x+5continue

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