We are given a function $f(x)$ defined as a determinant: $f(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix}$ We need to find the value of $f''(x) + f(x)$.

1. Problem Description

We are given a function $f(x)$ defined as a determinant:

$f(x) = \begin{vmatrix}

\sin x & \cos x & \sin x + \cos x + 1 \\

27 & 28 & 27 \\

1 & 1 & 1

\end{vmatrix}$

We need to find the value of $f''(x) + f(x)$.

2. Solution Steps

First, we evaluate the determinant:

$f(x) = \sin x (28 - 27) - \cos x (27 - 27) + (\sin x + \cos x + 1) (27 - 28)$

$f(x) = \sin x (1) - \cos x (0) + (\sin x + \cos x + 1) (-1)$

$f(x) = \sin x - \sin x - \cos x - 1$

$f(x) = -\cos x - 1$

Now, we find the first derivative of $f(x)$:

$f'(x) = \frac{d}{dx}(-\cos x - 1) = \sin x$

Next, we find the second derivative of $f(x)$:

$f''(x) = \frac{d}{dx}(\sin x) = \cos x$

Finally, we calculate $f''(x) + f(x)$:

$f''(x) + f(x) = \cos x + (-\cos x - 1) = \cos x - \cos x - 1 = -1$

3. Final Answer

The value of $f''(x) + f(x)$ is $-1$.