If x_0 is the point of local minima of f(x) = | vedclass

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(B) $f(x) = \bar{a} \cdot (\bar{b} 	imes \bar{c}) = \begin{vmatrix} x & -2 & 3 \ -2 & x & -1 \ 7 & -2 & x \end{vmatrix}$ Expanding the determinant:

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$-15$

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(B) $f(x) = \bar{a} \cdot (\bar{b} 	imes \bar{c}) = \begin{vmatrix} x & -2 & 3 \ -2 & x & -1 \ 7 & -2 & x \end{vmatrix}$ Expanding the determinant: $f(x) = x(x^2 - 2) + 2(-2x + 7) + 3(4 - 7x)$ $f(x) = x^3 - 2x - 4x + 14 + 12 - 21x$ $f(x) = x^3 - 27x + 26$ To find local minima,we find $f'(x) = 0$: $f'(x) = 3x^2 - 27 = 0$ $3(x^2 - 9) = 0 \Rightarrow x = \pm 3$ Using the second derivative test: $f''(x) = 6x$ $f''(3) = 18 > 0$,so $x_0 = 3$ is the point of local minima. At $x = 3$: $\bar{a} = 3 \hat{i} - 2 \hat{j} + 3 \hat{k}$ $\bar{b} = -2 \hat{i} + 3 \hat{j} - \hat{k}$ $\bar{a} \cdot \bar{b} = (3)(-2) + (-2)(3) + (3)(-1)$ $\bar{a} \cdot \bar{b} = -6 - 6 - 3 = -15$

If $x_0$ is the point of local minima of $f(x) = \bar{a} \cdot (\bar{b} \times \bar{c})$ where $\bar{a} = x \hat{i} - 2 \hat{j} + 3 \hat{k}$,$\bar{b} = -2 \hat{i} + x \hat{j} - \hat{k}$,$\bar{c} = 7 \hat{i} - 2 \hat{j} + x \hat{k}$,then the value of $\bar{a} \cdot \bar{b}$ at $x = x_0$ is

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