Let $$A = [a_{ij}]$$ be a $$3 \times 3$$ matrix, where $$a_{ij} = \begin{cases} 1, & \text{if } i = j \\ -x, & \text{if } |i-j| = 1 \\ 2x+1, & \text{otherwise} \end{cases}$$
Let a function $$f : R \to R$$ be defined as $$f(x) = \det(A)$$. Then the sum of maximum and minimum values of $$f$$ on $$R$$ is equal to:

The matrix is: $$A = \begin{bmatrix} 1 & -x & 2x+1 \\ -x & 1 & -x \\ 2x+1 & -x & 1 \end{bmatrix}.$$

Expanding the determinant along the first row: $$f(x) = \det(A) = 1(1 - x^2) - (-x)(-x - (-x)(2x+1)) + (2x+1)(x^2 - (2x+1)).$$

The second term: $$-(-x)(-x - (-x)(2x+1)) = x(-x + x(2x+1)) = x(-x + 2x^2 + x) = x(2x^2) = 2x^3$$.

The third term: $$(2x+1)(x^2 - 2x - 1) = 2x^3 - 4x^2 - 2x + x^2 - 2x - 1 = 2x^3 - 3x^2 - 4x - 1$$.

Adding all parts: $$f(x) = (1 - x^2) + 2x^3 + (2x^3 - 3x^2 - 4x - 1) = 4x^3 - 4x^2 - 4x.$$

To find critical points: $$f'(x) = 12x^2 - 8x - 4 = 4(3x^2 - 2x - 1) = 4(3x+1)(x-1) = 0$$, giving $$x = -\frac{1}{3}$$ or $$x = 1$$.

Using the second derivative $$f''(x) = 24x - 8$$:

At $$x = -\frac{1}{3}$$: $$f''\!\left(-\tfrac{1}{3}\right) = -16 < 0$$, so this is a local maximum. $$f\!\left(-\tfrac{1}{3}\right) = 4\!\left(-\tfrac{1}{27}\right) - 4\!\left(\tfrac{1}{9}\right) - 4\!\left(-\tfrac{1}{3}\right) = -\tfrac{4}{27} - \tfrac{12}{27} + \tfrac{36}{27} = \tfrac{20}{27}.$$

At $$x = 1$$: $$f''(1) = 16 > 0$$, so this is a local minimum. $$f(1) = 4 - 4 - 4 = -4$$.

The sum of the local maximum and local minimum values is $$\dfrac{20}{27} + (-4) = \dfrac{20}{27} - \dfrac{108}{27} = -\dfrac{88}{27}$$.