If m and M are respectively the absolute mini | vedclass

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(A) Given the function $f(x)=2x^3+9x^2+12x+1$ on the interval $[-3,0]$.First,find the derivative $f'(x)$:$f'(x)=6x^2+18x+12=6(x^2+3x+2)=6(x+1)(x+2)$.S

Vedclass · Accepted Answer

$-7$

Vedclass · Answer

(A) Given the function $f(x)=2x^3+9x^2+12x+1$ on the interval $[-3,0]$.First,find the derivative $f'(x)$:$f'(x)=6x^2+18x+12=6(x^2+3x+2)=6(x+1)(x+2)$.Setting $f'(x)=0$,we get critical points $x=-1$ and $x=-2$.Now,evaluate $f(x)$ at the critical points and the endpoints of the interval $[-3,0]$:$f(-3)=2(-27)+9(9)+12(-3)+1 = -54+81-36+1 = -8$.$f(-2)=2(-8)+9(4)+12(-2)+1 = -16+36-24+1 = -3$.$f(-1)=2(-1)+9(1)+12(-1)+1 = -2+9-12+1 = -4$.$f(0)=2(0)+9(0)+12(0)+1 = 1$.Comparing these values,the absolute minimum $m = -8$ and the absolute maximum $M = 1$.Therefore,$m+M = -8+1 = -7$.

If $m$ and $M$ are respectively the absolute minimum and absolute maximum values of a function $f(x)=2x^3+9x^2+12x+1$ defined on $[-3,0]$,then $m+M=$

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