Find the difference quotient of \( f \), that is find \( \frac{f(x+h)-f(x)}{h}, h \neq 0 \), for the function \( f(x)=\sqrt{x-19} \). [Hint: Rationalize the numerator.] The difference quotient of \( f ; f(x)=\sqrt{x-19} \) is \( \square \). (Simplify your answer.)

To find the difference quotient of the function \( f(x) = \sqrt{x - 19} \), we need to compute \( \frac{f(x+h) - f(x)}{h} \). 1. First, we find \( f(x + h) \): \[ f(x + h) = \sqrt{(x + h) - 19} = \sqrt{x + h - 19} = \sqrt{x - 19 + h}. \] 2. Now, substitute into the difference quotient: \[ \frac{f(x+h) - f(x)}{h} = \frac{\sqrt{x - 19 + h} - \sqrt{x - 19}}{h}. \] 3. To rationalize the numerator, we can multiply and divide by the conjugate: \[ = \frac{\left(\sqrt{x - 19 + h} - \sqrt{x - 19}\right) \left(\sqrt{x - 19 + h} + \sqrt{x - 19}\right)}{h\left(\sqrt{x - 19 + h} + \sqrt{x - 19}\right)}. \] The numerator simplifies using the difference of squares: \[ = \frac{(x - 19 + h) - (x - 19)}{h(\sqrt{x - 19 + h} + \sqrt{x - 19})}. \] 4. Simplifying the numerator: \[ = \frac{h}{h(\sqrt{x - 19 + h} + \sqrt{x - 19})}. \] 5. We can cancel \( h \) in the numerator and denominator: \[ = \frac{1}{\sqrt{x - 19 + h} + \sqrt{x - 19}}. \] Thus, the difference quotient of \( f \) is: \[ \frac{1}{\sqrt{x - 19 + h} + \sqrt{x - 19}}. \]