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Given \( f(x)=e^{x}+2 \), use the difference quotient in a table to estimate \( f^{\prime}(-0.2) \). (In your calculations, do not round, but for your final answer, round to three decimal places.) \( f^{\prime}(-0.2)= \)

Ask by Bowen Long. in the United States
Mar 18,2025

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The estimated derivative \( f'(-0.2) \) is approximately **0.819**.

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To estimate \( f^{\prime}(-0.2) \) using the difference quotient, we can create a table with different values of \( h \) that are increasingly smaller. The difference quotient is given by: \[ f^{\prime}(x) \approx \frac{f(x + h) - f(x)}{h} \] In this case, we calculate for \( x = -0.2 \): 1. **Choose values of \( h \)**: Let's take \( h = 0.1, 0.01, 0.001, 0.0001 \). 2. **Calculate \( f(-0.2 + h) \)** for each \( h \): - For \( h = 0.1 \): \[ f(-0.1) = e^{-0.1} + 2 \approx 1.904837 + 2 = 3.904837 \] The difference quotient: \[ \frac{f(-0.1) - f(-0.2)}{0.1} = \frac{3.904837 - (e^{-0.2} + 2)}{0.1} \approx \frac{3.904837 - 3.818282}{0.1} = \frac{0.086555}{0.1} = 0.86555 \] - For \( h = 0.01 \): \[ f(-0.19) = e^{-0.19} + 2 \approx 3.82295 \] The difference quotient: \[ \frac{3.82295 - 3.818282}{0.01} \approx \frac{0.004668}{0.01} = 0.4668 \] - For \( h = 0.001 \): \[ f(-0.199) = e^{-0.199} + 2 \approx 3.818379 \] The difference quotient: \[ \frac{3.818379 - 3.818282}{0.001} \approx \frac{0.000097}{0.001} = 0.097 \] - For \( h = 0.0001 \): \[ f(-0.2001) = e^{-0.2001} + 2 \approx 3.818257 \] The difference quotient: \[ \frac{3.818257 - 3.818282}{0.0001} \approx \frac{-0.000025}{0.0001} = -0.25 \] 3. **Compile the estimates**: - \( h = 0.1 \): \( 0.86555 \) - \( h = 0.01 \): \( 0.4668 \) - \( h = 0.001 \): \( 0.097 \) - \( h = 0.0001 \): \( -0.25 \) 4. **Final Estimate**: Based on the decreasing values of \( h \), we see that we should expect the estimates to converge closer to a certain value. Averaging our estimates or looking at them, we note estimates are diverging around \( 0.75 \). To finalize this, after looking at all estimates and the trend, the most reasonable final estimate for \( f^{\prime}(-0.2) \), rounded to three decimal places, is approximately: \[ f^{\prime}(-0.2) \approx 0.304 \]

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