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Question Number 209332 by efronzo1 last updated on 07/Jul/24

Answered by Frix last updated on 07/Jul/24

x^3 −8x^2 +(16−k)x=0 x_1 =0 x_2 =4−(√k) x_3 =4+(√k) ∫_0 ^(4+(√k)) (x^3 −8x^2 +(16−k)x)dx=0 −(((4+(√k))^2 (3k+8(√k)−16))/(12))=0 k=((16)/9)

x3−8x2+(16−k)x=0x1=0x2=4−kx3=4+k∫4+k0(x3−8x2+(16−k)x)dx=0−(4+k)2(3k+8k−16)12=0k=169

Answered by MM42 last updated on 07/Jul/24

f(x)=x^3 −8x^2 +16x ⇒A((8/3) , ((128)/(27))) : symmetry center ((128)/(27))=k×(8/3)⇒k=((16)/9) ✓

f(x)=x3−8x2+16x⇒A(83,12827):symmetrycenter12827=k×83⇒k=169✓

Answered by efronzo1 last updated on 08/Jul/24

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