Find the real number $k$ such that $f(x) = f(k - x)$ for all real numbers $x$, where

To find the value of \( k \) such that \( f(x) = f(k - x) \) for all \( x \), we first need to simplify the function \( f(x) \).

The function \( f(x) \) is given by:

\[
f(x) = 3x^2 - 4x - x^2 + 11x.
\]

We combine like terms:
\[
f(x) = (3x^2 - x^2) + (-4x + 11x) = 2x^2 + 7x.
\]

Next, we will compute \( f(k - x) \):
\[
f(k - x) = 2(k - x)^2 + 7(k - x).
\]
Expanding \( (k - x)^2 \):
\[
(k - x)^2 = k^2 - 2kx + x^2.
\]
Thus,
\[
f(k - x) = 2(k^2 - 2kx + x^2) + 7(k - x).
\]
Distributing gives:
\[
f(k - x) = 2k^2 - 4kx + 2x^2 + 7k - 7x.
\]

Combining the terms, we have:
\[
f(k - x) = 2x^2 + (-4k - 7)x + (2k^2 + 7k).
\]

Now we set \( f(x) \) equal to \( f(k - x) \):
\[
2x^2 + 7x = 2x^2 + (-4k - 7)x + (2k^2 + 7k).
\]

Since \( 2x^2 \) appears on both sides of the equation, we can simplify to:
\[
7x = (-4k - 7)x + (2k^2 + 7k).
\]

To satisfy this equation for all \( x \), the coefficients of \( x \) must be equal and the constant terms must be equal. First, we set the coefficients of \( x \):
\[
7 = -4k - 7.
\]
Solving for \( k \):
\[
7 + 7 = -4k \implies 14 = -4k \implies k = -\frac{14}{4} = -\frac{7}{2}.
\]

Next, we check if the constant terms match. The left-hand side has no constant term (implicitly 0), while the right-hand side constant term is:
\[
2k^2 + 7k.
\]
Substituting \( k = -\frac{7}{2} \):
\[
2\left(-\frac{7}{2}\right)^2 + 7\left(-\frac{7}{2}\right) = 2\cdot\frac{49}{4} - \frac{49}{2} = \frac{98}{4} - \frac{98}{4} = 0.
\]

Since both conditions are satisfied, the value of \( k \) is:
\[
\boxed{-\frac{7}{2}}.
\]