Key Concepts and Formulas

• Determinant of a 3x3 Matrix: For a matrix $A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$, the determinant is given by $det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$.
• Finding Critical Points: Critical points of a function $f(x)$ occur where its derivative $f'(x) = 0$ or is undefined.
• Second Derivative Test: If $f'(c) = 0$, then:
  • If $f''(c) > 0$, then $f(x)$ has a local minimum at $x = c$.
  • If $f''(c) < 0$, then $f(x)$ has a local maximum at $x = c$.

Step-by-Step Solution

Step 1: Construct the Matrix A

We are given the rule for the elements $a_{ij}$ of the 3x3 matrix A: a_{ij} = \left\{ {\begin{array}{*{20}{l}} 1 & , & {if\,i = j} \\ { - x} & , & {if\,\left| {i - j} \right| = 1} \\ {2x + 1} & , & {otherwise.} \end{array}} \right. This means:

• Diagonal elements ($i = j$) are 1.
• Elements adjacent to the diagonal ($|i - j| = 1$) are $-x$.
• All other elements are $2x + 1$.

Therefore, the matrix A is: $A = \begin{bmatrix} 1 & -x & 2x+1 \\ -x & 1 & -x \\ 2x+1 & -x & 1 \end{bmatrix}$ We construct the matrix A according to the given conditions for its elements.

Step 2: Calculate the Determinant of A (f(x))

The function $f(x)$ is defined as the determinant of A. We compute the determinant using the formula: $det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$ Substituting the elements of A: $f(x) = 1(1 \cdot 1 - (-x)(-x)) - (-x)((-x) \cdot 1 - (-x)(2x+1)) + (2x+1)((-x)(-x) - 1(2x+1))$ $f(x) = 1(1 - x^2) + x(-x + 2x^2 + x) + (2x+1)(x^2 - 2x - 1)$ $f(x) = 1 - x^2 + 2x^3 + (2x^3 - 4x^2 - 2x + x^2 - 2x - 1)$ $f(x) = 1 - x^2 + 2x^3 + 2x^3 - 3x^2 - 4x - 1$ $f(x) = 4x^3 - 4x^2 - 4x$ We calculate the determinant to obtain an explicit expression for $f(x)$.

Step 3: Find the First Derivative f'(x)

To find the critical points, we need the first derivative of $f(x)$: $f(x) = 4x^3 - 4x^2 - 4x$ $f'(x) = 12x^2 - 8x - 4$ We differentiate $f(x)$ to find potential locations of maxima and minima.

Step 4: Find the Critical Points

Set $f'(x) = 0$ and solve for $x$: $12x^2 - 8x - 4 = 0$ Divide by 4: $3x^2 - 2x - 1 = 0$ Factor the quadratic: $(3x + 1)(x - 1) = 0$ The critical points are: $x = -\frac{1}{3}, \quad x = 1$ The roots of $f'(x) = 0$ are the critical points where local extrema can exist.

Step 5: Find the Second Derivative f''(x)

To classify the critical points, we need the second derivative: $f'(x) = 12x^2 - 8x - 4$ $f''(x) = 24x - 8$ We differentiate $f'(x)$ to determine the concavity of $f(x)$ at the critical points.

Step 6: Classify the Critical Points using the Second Derivative Test

Evaluate $f''(x)$ at each critical point:

• For $x = -\frac{1}{3}$: $f''\left(-\frac{1}{3}\right) = 24\left(-\frac{1}{3}\right) - 8 = -8 - 8 = -16 < 0$ Since $f''(-\frac{1}{3}) < 0$, there is a local maximum at $x = -\frac{1}{3}$.
• For $x = 1$: $f''(1) = 24(1) - 8 = 16 > 0$ Since $f''(1) > 0$, there is a local minimum at $x = 1$. The sign of $f''(x)$ at the critical points tells us whether they are local maxima or minima.

Step 7: Calculate the Maximum and Minimum Values

Evaluate $f(x)$ at the critical points:

• For $x = -\frac{1}{3}$: $f\left(-\frac{1}{3}\right) = 4\left(-\frac{1}{3}\right)^3 - 4\left(-\frac{1}{3}\right)^2 - 4\left(-\frac{1}{3}\right) = 4\left(-\frac{1}{27}\right) - 4\left(\frac{1}{9}\right) + \frac{4}{3} = -\frac{4}{27} - \frac{12}{27} + \frac{36}{27} = \frac{20}{27}$ So, the local maximum value is $\frac{20}{27}$.
• For $x = 1$: $f(1) = 4(1)^3 - 4(1)^2 - 4(1) = 4 - 4 - 4 = -4$ So, the local minimum value is $-4$. We plug the critical points into $f(x)$ to find the maximum and minimum function values.

Step 8: Find the Sum of the Maximum and Minimum Values

The sum of the maximum and minimum values is: $\frac{20}{27} + (-4) = \frac{20}{27} - \frac{108}{27} = -\frac{88}{27}$ We add the maximum and minimum values as requested.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when calculating the determinant and derivatives. A small sign error can propagate through the entire solution.
• Factoring Quadratics: Practice factoring quadratic equations quickly and accurately. Use the quadratic formula if necessary.
• Second Derivative Test: Remember that $f''(x) > 0$ implies a local minimum, and $f''(x) < 0$ implies a local maximum.

Summary

We constructed the matrix $A$ based on the given rules, calculated its determinant to obtain the function $f(x) = 4x^3 - 4x^2 - 4x$. We then found the critical points by setting $f'(x) = 0$, classified them using the second derivative test, and calculated the corresponding maximum and minimum values. Finally, we summed these values to obtain the answer.

Final Answer

The final answer is \boxed{-{{88} \over {27}}}, which corresponds to option (D).