## Calculus (3rd Edition)

The mean of these two symmetric difference quotients is $-0.39375 \mathrm{kph}\times\mathrm{km} / \mathrm{car}$.
Let $S(q)$ be the function determining $S$ given $q$. The symmetric difference quotient with $h=10$ is $$S^{\prime}(80) \approx \frac{S(90)-S(70)}{20}=\frac{60-67.5}{20}=-0.375$$ with $h=20$ $$S^{\prime}(80) \approx \frac{S(100)-S(60)}{40}=\frac{56-72.5}{40}=-0.4125$$ The mean of these two symmetric difference quotients is $-0.39375 \mathrm{kph}\times\mathrm{km} / \mathrm{car}$.