Functional equation: \[ f(x + y) = f(x)f(y) \] and \[ f(5) = 4 \]
Substitute \( y = 0 \) into the functional equation: \[ f(x + 0) = f(x)f(0) \Rightarrow f(x) = f(x)f(0) \] Assuming \( f(x) eq 0 \), divide both sides by \( f(x) \): \[ f(0) = 1 \]
\[ f(10) = f(5 + 5) = f(5) \cdot f(5) = 4 \cdot 4 = 16 \]
Utilize the identity \( f(0) = 1 \): \[ f(0) = f(5 + (-5)) = f(5)f(-5) \Rightarrow 1 = 4 \cdot f(-5) \] Therefore: \[ f(-5) = \frac{1}{4} \]
\[ f(-10) = f(-5 + (-5)) = f(-5) \cdot f(-5) = \left( \frac{1}{4} \right)^2 = \frac{1}{16} \]
\[ f(10) - f(-10) = 16 - \frac{1}{16} = \frac{256 - 1}{16} = \frac{255}{16} = \boxed{15.9375} \]
\(\boxed{15.9375}\)