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(C) First,find the critical points of $f(x)=2 x^{3}-3 x^{2}-12 x$ by setting $f^{\prime}(x)=0$.$f^{\prime}(x)=6 x^{2}-6 x-12=6(x-2)(x+1)$.The critical

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$114$

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(C) First,find the critical points of $f(x)=2 x^{3}-3 x^{2}-12 x$ by setting $f^{\prime}(x)=0$.$f^{\prime}(x)=6 x^{2}-6 x-12=6(x-2)(x+1)$.The critical points are $x=-1$ and $x=2$.Using the second derivative test,$f^{\prime\prime}(x)=12 x-6$.$f^{\prime\prime}(-1)=-18 $f^{\prime\prime}(2)=18 > 0$,so $b=2$ is the point of local minimum.The area $A$ is given by $\int_{a}^{b} |f(x)| dx = \int_{-1}^{2} |2 x^{3}-3 x^{2}-12 x| dx$.The function $f(x)=x(2 x^{2}-3 x-12)$ crosses the $x$-axis at $x=0$ within the interval $[-1, 2]$.Thus,$A = \int_{-1}^{0} (2 x^{3}-3 x^{2}-12 x) dx + \int_{0}^{2} -(2 x^{3}-3 x^{2}-12 x) dx$.$A = \left[ \frac{x^{4}}{2} - x^{3} - 6 x^{2} ight]_{-1}^{0} + \left[ 6 x^{2} + x^{3} - \frac{x^{4}}{2} ight]_{0}^{2}$.$A = (0 - (\frac{1}{2} + 1 - 6)) + ((24 + 8 - 8) - 0) = -(\frac{1-10}{2}) + 24 = \frac{9}{2} + 24 = \frac{57}{2}$.Therefore,$4 A = 4 	imes \frac{57}{2} = 114$.

Let $a$ and $b$ respectively be the points of local maximum and local minimum of the function $f(x)=2 x^{3}-3 x^{2}-12 x$. If $A$ is the total area of the region bounded by $y=f(x)$,the $x$-axis and the lines $x=a$ and $x=b$,then $4 A$ is equal to ..... .

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