In Exercises 5-8, \(f(x)\) is the number of meters a bus has gone at a time \(x\) (in seconds). Find the instantaneous velocity at the given time \(x_0\).
\begin{align*} \frac{\Delta y}{\Delta x} & = \frac{f(2 + \Delta x) - f(2)}{\Delta x} \\ & = \frac{(10 + 7\Delta x + (\Delta x)^2) - 10}{\Delta x} \\ & = \frac{7\Delta x + (\Delta x)^2}{\Delta x} \\ & = 7 + \Delta x \end{align*} Let \(\Delta x\) become very small. The required instantaneous velocity at \(x_0 = 2\) is 7 meters per second.
\( \Delta y = f(2 + 0.5) - f(2) = 7.25\). Avg. Velocity = \( \frac{\Delta y}{\Delta x} = \frac{7.25}{0.5} = 14.5 \)
\( \Delta y = f(2 + 0.01) - f(2) = 0.1303 \). Avg. Velocity = \( \frac{\Delta y}{\Delta x} = \frac{0.1303}{0.01} = 13.01 \)
\( \Delta y = f(4 + 0.1) - f(4) = 2.53 \). Avg. Velocity = \( \frac{\Delta y}{\Delta x} = \frac{2.53}{0.1} = 25.3 \)
\( \Delta y = f(4 + 0.01) - f(4) = 0.2503 \). Avg. Velocity = \( \frac{\Delta y}{\Delta x} = \frac{0.2503}{0.01} = 25.03 \)
\( \Delta y = f(2 + 0.5) - f(2) = 7.25\). Avg. Velocity = \( \frac{\Delta y}{\Delta x} = \frac{7.25}{0.5} = 14.5 \)
\( \Delta y = f(2 + 0.01) - f(2) = 0.1401 \). Avg. Velocity = \( \frac{\Delta y}{\Delta x} = \frac{0.1401}{0.01} = 14.01 \)
\( \Delta y = f(4 + 0.1) - f(4) = 1.81 \). Avg. Velocity = \( \frac{\Delta y}{\Delta x} = \frac{1.81}{0.1} = 18.1 \)
\( \Delta y = f(4 + 0.01) - f(4) = 0.1801 \). Avg. Velocity = \( \frac{\Delta y}{\Delta x} = \frac{0.2503}{0.01} = 18.01 \)
\( \Delta y = f(2 + 0.5) - f(2) = 1\). Avg. Velocity = \( \frac{\Delta y}{\Delta x} = \frac{1}{0.5} = 0.5 \)
\( \Delta y = f(2 + 0.01) - f(2) = 0.02 \). Avg. Velocity = \( \frac{\Delta y}{\Delta x} = \frac{0.02}{0.01} = 2 \)
\( \Delta y = f(4 + 0.1) - f(4) = .2 \). Avg. Velocity = \( \frac{\Delta y}{\Delta x} = \frac{.2}{0.1} = 2 \)/p>
\( \Delta y = f(4 + 0.01) - f(4) = 0.02 \). Avg. Velocity = \( \frac{\Delta y}{\Delta x} = \frac{0.02}{0.01} = 2 \)