Let x_0 be the point of local minima of f(x) | vedclass

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(B) The function $f(x)$ is defined as the scalar triple product of vectors $\overline{a}, \overline{b}, 	ext{ and } \overline{c}$:$f(x) = \begin{vmat

Vedclass · Accepted Answer

$-15$

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(B) The function $f(x)$ is defined as the scalar triple product of vectors $\overline{a}, \overline{b}, 	ext{ and } \overline{c}$:$f(x) = \begin{vmatrix} x & -2 & 3 \ -2 & x & -1 \ 7 & -2 & x \end{vmatrix}$Expanding the determinant along the first row:$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 - 19x - 2$ (Note: Correcting the expansion: $x(x^2-2) + 2(-2x+7) + 3(4-7x) = x^3 - 2x - 4x + 14 + 12 - 21x = x^3 - 27x + 26$)To find the local minima,we find the critical points by setting $f'(x) = 0$:$f'(x) = 3x^2 - 27 = 0$$3(x^2 - 9) = 0 \Rightarrow x = \pm 3$Using the second derivative test,$f''(x) = 6x$:At $x = 3$,$f''(3) = 18 > 0$,so $x_0 = 3$ is the point of local minima.At $x = 3$,$\overline{a} = 3 \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $\overline{b} = -2 \hat{i} + 3 \hat{j} - \hat{k}$.Calculating the dot product $\overline{a} \cdot \overline{b}$:$\overline{a} \cdot \overline{b} = (3)(-2) + (-2)(3) + (3)(-1) = -6 - 6 - 3 = -15$.

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