`a_(ij)={(1,,i=j),(-x,,abs(i-j)=1),(2x+1,,otherwise):} , A={a_(ij)}_(3xx3) f(x)=det(A)`. then sum of maximum and minimum vaues of f(x) is

Answer

To solve the problem, we start by constructing the matrix \( A \) based on the given conditions for \( a_{ij} \):

1. **Matrix Construction**:
- For \( i = j \) (diagonal elements), \( a_{ij} = 1 \).
- For \( |i - j| = 1 \) (adjacent elements), \( a_{ij} = -x \).
- For all other cases, \( a_{ij} = 2x + 1 \).

Thus, the \( 3 \times 3 \) matrix \( A \) can be constructed as follows:

\[ A = \begin{pmatrix} 1 & -x & 2x + 1 \\ -x & 1 & -x \\ 2x + 1 & -x & 1 \end{pmatrix} \]

2. **Finding the Determinant**:
We need to find the determinant of matrix \( A \), denoted as \( f(x) = \det(A) \).

Using the determinant formula for a \( 3 \times 3 \) matrix:

\[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

where \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \).

For our matrix \( A \):

- \( a = 1, b = -x, c = 2x + 1 \)
- \( d = -x, e = 1, f = -x \)
- \( g = 2x + 1, h = -x, i = 1 \)

Now substituting these values into the determinant formula:

\[ f(x) = 1 \cdot (1 \cdot 1 - (-x)(-x)) - (-x) \cdot (-x \cdot 1 - (-x)(2x + 1)) + (2x + 1) \cdot (-x \cdot -x - 1 \cdot (2x + 1)) \]

Calculating each term step-by-step:

- First term: \( 1 \cdot (1 - x^2) = 1 - x^2 \)
- Second term: \( -(-x) \cdot (-x - (-x)(2x + 1)) = x \cdot (-x + 2x^2 + x) = x(2x^2) = 2x^3 \)
- Third term: \( (2x + 1) \cdot (x^2 - (2x + 1)) = (2x + 1)(x^2 - 2x - 1) \)

Now simplifying the third term:

\[ (2x + 1)(x^2 - 2x - 1) = 2x^3 - 4x^2 - 2x + x^2 - 2x - 1 = 2x^3 - 3x^2 - 4x - 1 \]

Combining all terms together:

\[ f(x) = (1 - x^2) + 2x^3 + (2x^3 - 3x^2 - 4x - 1) \]

This simplifies to:

\[ f(x) = 4x^3 - 4x^2 - 4x \]

3. **Finding Maximum and Minimum Values**:
To find the maximum and minimum values of \( f(x) \), we differentiate \( f(x) \):

\[ f'(x) = 12x^2 - 8x - 4 \]

Setting the derivative equal to zero to find critical points:

\[ 12x^2 - 8x - 4 = 0 \]

Dividing through by 4:

\[ 3x^2 - 2x - 1 = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm 4}{6} \]

This gives us:

\[ x = 1 \quad \text{and} \quad x = -\frac{1}{3} \]

4. **Evaluating \( f(x) \) at Critical Points**:
Now we evaluate \( f(x) \) at these critical points:

- For \( x = 1 \):

\[ f(1) = 4(1)^3 - 4(1)^2 - 4(1) = 4 - 4 - 4 = -4 \]

- For \( x = -\frac{1}{3} \):

\[ f\left(-\frac{1}{3}\right) = 4\left(-\frac{1}{3}\right)^3 - 4\left(-\frac{1}{3}\right)^2 - 4\left(-\frac{1}{3}\right) \]

Calculating each term:

\[ = 4\left(-\frac{1}{27}\right) - 4\left(\frac{1}{9}\right) + \frac{4}{3} \]
\[ = -\frac{4}{27} - \frac{12}{27} + \frac{36}{27} = \frac{20}{27} \]

5. **Finding the Sum of Maximum and Minimum Values**:
Now we find the maximum and minimum values:

- Minimum value: \( -4 \)
- Maximum value: \( \frac{20}{27} \)

Thus, the sum of the maximum and minimum values is:

\[ \text{Sum} = -4 + \frac{20}{27} = -\frac{108}{27} + \frac{20}{27} = -\frac{88}{27} \]

**Final Answer**:
The sum of the maximum and minimum values of \( f(x) \) is \( -\frac{88}{27} \).

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