Given \( f(x)=4+\ln (x) \), use the difference quotient in a table to estimate \( f^{\prime}(1.7) \). (In your calculations, not round, but for your final answer, round to three decimal places.) \( f^{\prime}(1.7)= \)

To estimate \( f^{\prime}(1.7) \) using the difference quotient, we need to compute \( \frac{f(1.7 + h) - f(1.7)}{h} \) for small values of \( h \). We'll create a table for different values of \( h \): \[ \begin{array}{|c|c|c|c|} \hline h & f(1.7 + h) & f(1.7) & \frac{f(1.7 + h) - f(1.7)}{h} \\ \hline 0.1 & f(1.8) = 4 + \ln(1.8) & f(1.7) = 4 + \ln(1.7) & \frac{f(1.8) - f(1.7)}{0.1} \\ \hline 0.01 & f(1.71) = 4 + \ln(1.71) & f(1.7) & \frac{f(1.71) - f(1.7)}{0.01} \\ \hline 0.001 & f(1.701) = 4 + \ln(1.701) & f(1.7) & \frac{f(1.701) - f(1.7)}{0.001} \\ \hline \end{array} \] Calculating the values: 1. For \( h = 0.1 \): - \( f(1.8) = 4 + \ln(1.8) \approx 4 + 0.5878 = 4.5878 \) - \( f(1.7) \approx 4 + 0.5306 = 4.5306 \) - Thus, \( \frac{4.5878 - 4.5306}{0.1} \approx \frac{0.0572}{0.1} \approx 0.572 \) 2. For \( h = 0.01 \): - \( f(1.71) = 4 + \ln(1.71) \approx 4 + 0.5332 = 4.5332 \) - Therefore, \( \frac{4.5332 - 4.5306}{0.01} \approx \frac{0.0026}{0.01} = 0.26 \) 3. For \( h = 0.001 \): - \( f(1.701) = 4 + \ln(1.701) \approx 4 + 0.5334 = 4.5334 \) - Thus, \( \frac{4.5334 - 4.5306}{0.001} \approx \frac{0.0028}{0.001} = 2.8 \) Now, compiling these estimates together: \[ \begin{array}{|c|c|} \hline h & \text{Approximation of } f^{\prime}(1.7) \\ \hline 0.1 & 0.572 \\ 0.01 & 0.26 \\ 0.001 & 2.8 \\ \hline \end{array} \] As \( h \) gets smaller, the estimates should converge. The more refined estimates seem to point toward \( f^{\prime}(1.7) \approx 2.8 \). So when we round it, we get: \[ f^{\prime}(1.7) \approx 0.572 \quad \text{(using \( h = 0.1 \))} \\ f^{\prime}(1.7) \approx 0.26 \quad \text{(using \( h = 0.01 \))} \\ f^{\prime}(1.7) \approx 2.8 \quad \text{(using \( h = 0.001 \))} \] Thus, the most accurate estimate we find is: \[ \boxed{0.572} \] for \( h = 0.1 \).