Find all twice differentiable functions $f(x)$ such that $f^{\prime \prime}(x)=0$, $f(0)=19$, and $f(1)=99$.
1. Given that \( f''(x) = 0 \), we start by integrating this equation with respect to \( x \):
   \[
   \int f''(x) \, dx = \int 0 \, dx \implies f'(x) = c_1
   \]
   where \( c_1 \) is a constant of integration.

2. Next, we integrate \( f'(x) = c_1 \) with respect to \( x \):
   \[
   \int f'(x) \, dx = \int c_1 \, dx \implies f(x) = c_1 x + c_2
   \]
   where \( c_2 \) is another constant of integration.

3. We now use the initial conditions to determine the constants \( c_1 \) and \( c_2 \):
   - From \( f(0) = 19 \):
     \[
     f(0) = c_1 \cdot 0 + c_2 = 19 \implies c_2 = 19
     \]
   - From \( f(1) = 99 \):
     \[
     f(1) = c_1 \cdot 1 + c_2 = 99 \implies c_1 + 19 = 99 \implies c_1 = 80
     \]

4. Substituting \( c_1 \) and \( c_2 \) back into the function \( f(x) \), we get:
   \[
   f(x) = 80x + 19
   \]

Thus, the function that satisfies all the given conditions is:
\[
\boxed{f(x) = 80x + 19}
\]