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

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(D) The function $f(x)$ is the scalar triple product $[\vec{a} \vec{b} \vec{c}]$, given by the determinant:$f(x) = \begin{vmatrix} x & -2 & 3 \ -2 &

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$-26$

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(D) The function $f(x)$ is the scalar triple product $[\vec{a} \vec{b} \vec{c}]$, given by the determinant:$f(x) = \begin{vmatrix} x & -2 & 3 \ -2 & x & -1 \ 7 & -2 & x \end{vmatrix}$Expanding 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 = x^3 - 27x + 26$To find local maxima, find $f'(x) = 3x^2 - 27$. Setting $f'(x) = 0$ gives $x^2 = 9$, so $x = \pm 3$.$f''(x) = 6x$. For local maxima, $f''(x) Now calculate $\vec{a} \cdot \vec{c}$ at $x = -3$:$\vec{a} = -3\hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{c} = 7\hat{i} - 2\hat{j} - 3\hat{k}$.$\vec{a} \cdot \vec{c} = (-3)(7) + (-2)(-2) + (3)(-3) = -21 + 4 - 9 = -26$.

Let $x_0$ be the point of local maxima of $f(x) = \vec{a} \cdot (\vec{b} \times \vec{c})$, where $\vec{a} = x\hat{i} - 2\hat{j} + 3\hat{k}$, $\vec{b} = -2\hat{i} + x\hat{j} - \hat{k}$, and $\vec{c} = 7\hat{i} - 2\hat{j} + x\hat{k}$. Then the value of $\vec{a} \cdot \vec{c}$ at $x = x_0$ is:

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