Let x_{0} be the point of local maxima of f(x | vedclass

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(D) The function is defined as the scalar triple product $f(x) = [\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} x & -2 & 3 \ -2 & x & -1 \ 7 & -2 & x \

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$-22$

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(D) The function is defined as the scalar triple product $f(x) = [\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} x & -2 & 3 \ -2 & x & -1 \ 7 & -2 & x \end{vmatrix}$.Expanding the determinant: $f(x) = x(x^2 - 2) + 2(-2x + 7) + 3(4 - 7x) = x^3 - 2x - 4x + 14 + 12 - 21x = x^3 - 27x + 26$.To find the local maxima,we calculate the derivative: $f'(x) = 3x^2 - 27$. Setting $f'(x) = 0$ gives $x^2 = 9$,so $x = \pm 3$.The second derivative is $f''(x) = 6x$. For $x = -3$,$f''(-3) = -18 At $x = -3$,the vectors are $\vec{a} = -3\hat{i} - 2\hat{j} + 3\hat{k}$,$\vec{b} = -2\hat{i} - 3\hat{j} - \hat{k}$,and $\vec{c} = 7\hat{i} - 2\hat{j} - 3\hat{k}$.Calculating the dot products:$\vec{a} \cdot \vec{b} = (-3)(-2) + (-2)(-3) + (3)(-1) = 6 + 6 - 3 = 9$.$\vec{b} \cdot \vec{c} = (-2)(7) + (-3)(-2) + (-1)(-3) = -14 + 6 + 3 = -5$.$\vec{c} \cdot \vec{a} = (7)(-3) + (-2)(-2) + (-3)(3) = -21 + 4 - 9 = -26$.Summing these: $9 - 5 - 26 = -22$.

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