Use \( f(x)=\sqrt{x} \) to approximate \( \sqrt{8} \) using a tangent line. \[ \sqrt{8} \approx \frac{[?]}{\square} \]

To approximate \( \sqrt{8} \) using a tangent line, we first need to find the derivative of the function \( f(x) = \sqrt{x} \). 1. The derivative is: \[ f'(x) = \frac{1}{2\sqrt{x}}. \] 2. We will use the point \( x = 9 \) (since \( 9 \) is close to \( 8 \) and is a perfect square). The value of the function and its derivative at this point is: \[ f(9) = \sqrt{9} = 3, \quad f'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{6}. \] 3. Now, applying the formula for the tangent line: \[ L(x) = f(9) + f'(9)(x - 9) = 3 + \frac{1}{6}(x - 9). \] 4. To approximate \( \sqrt{8} \), substitute \( x = 8 \): \[ L(8) = 3 + \frac{1}{6}(8 - 9) = 3 - \frac{1}{6} = \frac{18}{6} - \frac{1}{6} = \frac{17}{6}. \] Thus, we can conclude that \( \sqrt{8} \approx \frac{17}{6} \).