Exercise 1.1.1-4.

In Exercises 1-4, $y$ represents the distance a bus has travelled after $x$ seconds. Find $Δ y$ and the average velocity during the interval $Δ x$ for the following situations.

$x_{0} = 2, Δ x = 0.5$
$x_{0} = 2, Δ x = 0.01$
$x_{0} = 4, Δ x = 0.1$
$x_{0} = 4, Δ x = 0.01$

Formulas:

$Δ y = f (x_{0} + Δ x) - f (x_{0})$

Average Velocity: $\frac{Δ y}{Δ x}$

Exercise 1. $y = x^{2} + 3 x$

Solution:

$Δ y = f (2 + 0.5) - f (2) = 3.75$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{3.75}{0.5} = 7.5$
$Δ y = f (2 + 0.01) - f (2) = 0.0701$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{0.0701}{0.01} = 7.01$
$Δ y = f (4 + 0.1) - f (4) = 1.11$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{1.11}{0.1} = 11.1$
$Δ y = f (4 + 0.01) - f (4) = 0.1101$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{0.1101}{0.01} = 11.01$

Exercise 2. $y = 3 x^{2} + x$

Solution:

$Δ y = f (2 + 0.5) - f (2) = 7.25$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{7.25}{0.5} = 14.5$
$Δ y = f (2 + 0.01) - f (2) = 0.1303$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{0.1303}{0.01} = 13.01$
$Δ y = f (4 + 0.1) - f (4) = 2.53$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{2.53}{0.1} = 25.3$
$Δ y = f (4 + 0.01) - f (4) = 0.2503$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{0.2503}{0.01} = 25.03$

Exercise 3. $y = x^{2} + 10 x$

Solution:

$Δ y = f (2 + 0.5) - f (2) = 7.25$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{7.25}{0.5} = 14.5$
$Δ y = f (2 + 0.01) - f (2) = 0.1401$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{0.1401}{0.01} = 14.01$
$Δ y = f (4 + 0.1) - f (4) = 1.81$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{1.81}{0.1} = 18.1$
$Δ y = f (4 + 0.01) - f (4) = 0.1801$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{0.2503}{0.01} = 18.01$

Exercise 4. $y = 2 x$

Solution:

$Δ y = f (2 + 0.5) - f (2) = 1$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{1}{0.5} = 0.5$
$Δ y = f (2 + 0.01) - f (2) = 0.02$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{0.02}{0.01} = 2$
$Δ y = f (4 + 0.1) - f (4) = .2$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{.2}{0.1} = 2$ /p>
$Δ y = f (4 + 0.01) - f (4) = 0.02$ . Avg. Velocity = $\frac{Δ y}{Δ x} = \frac{0.02}{0.01} = 2$

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Exercise 1.1.1-4.

Exercise 1. y=x2+3xy=x2+3x

Solution:

Exercise 2. y=3x2+xy=3x2+x

Solution:

Exercise 3. y=x2+10xy=x2+10x

Solution:

Exercise 4. y=2xy=2x

Solution:

Exercise 1. $y = x^{2} + 3 x$

Exercise 2. $y = 3 x^{2} + x$

Exercise 3. $y = x^{2} + 10 x$

Exercise 4. $y = 2 x$