Question

a function is shown in the table below on which interval of x is the average rate of change of the function the GREATEST
left and right are side by side
x
1.0
2.2
3.11
4.26
5.32
y
1.3
2.12
3.24
4.36
5.40
a x=0 to x=2
bx=2 to x=11
c x=11 to x=26
d x=26 to x=32

Answer 1

To find the interval where the average rate of change of the function is the greatest, we first need to calculate the average rate of change for each of the given intervals. The average rate of change between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) can be calculated using the formula:

\[
\text{Average Rate of Change} = \frac{y_2 - y_1}{x_2 - x_1}
\]

Let's calculate the average rate of change for each of the intervals in the given options.

**1. Interval \( x=0 \) to \( x=2 \):**

Using points \( (1.0, 1.3) \) and \( (2.2, 2.12) \):
\[
\text{Average Rate of Change} = \frac{2.12 - 1.3}{2.2 - 1.0} = \frac{0.82}{1.2} \approx 0.6833
\]

**2. Interval \( x=2 \) to \( x=11 \):**

Using points \( (2.2, 2.12) \) and \( (3.11, 3.24) \):
\[
\text{Average Rate of Change} = \frac{3.24 - 2.12}{3.11 - 2.2} = \frac{1.12}{0.91} \approx 1.2308
\]

**3. Interval \( x=11 \) to \( x=26 \):**

Using points \( (3.11, 3.24) \) and \( (4.26, 4.36) \):
\[
\text{Average Rate of Change} = \frac{4.36 - 3.24}{4.26 - 3.11} = \frac{1.12}{1.15} \approx 0.9739
\]

**4. Interval \( x=26 \) to \( x=32 \):**

Using points \( (4.26, 4.36) \) and \( (5.32, 5.40) \):
\[
\text{Average Rate of Change} = \frac{5.40 - 4.36}{5.32 - 4.26} = \frac{1.04}{1.06} \approx 0.9811
\]

Now we compare the average rates of change:

- Interval \( x=0 \) to \( x=2 \): \( \approx 0.6833 \)
- Interval \( x=2 \) to \( x=11 \): \( \approx 1.2308 \)
- Interval \( x=11 \) to \( x=26 \): \( \approx 0.9739 \)
- Interval \( x=26 \) to \( x=32 \): \( \approx 0.9811 \)

The greatest average rate of change occurs in the interval \( x=2 \) to \( x=11 \).

**So the answer is: b) \( x=2 \) to \( x=11 \)**.