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To solve this problem, we need to calculate the average rate of change of the function \( f(x) = \frac{x^2}{8} \) over the interval \([a, a + h]\) with \( a = 1 \) and different values of \( h \). The average rate of change of a function \( f \) over an interval \([a, b]\) is given by:

\[
\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}
\]

Here, \( b = a + h \), so we substitute:

\[
\text{Average Rate of Change} = \frac{f(a + h) - f(a)}{h}
\]

Since \( a = 1 \):

\[
f(a) = f(1) = \frac{1^2}{8} = \frac{1}{8}
\]

Now, for each value of \( h \), we calculate \( f(1 + h) = \frac{(1 + h)^2}{8} \) and then plug into the average rate of change formula. Let's calculate these values:

1. For \( h = 1 \):

\[
f(1 + 1) = f(2) = \frac{2^2}{8} = \frac{4}{8} = 0.5
\]

\[
\text{Average Rate of Change} = \frac{0.5 - \frac{1}{8}}{1} = 0.5 - 0.125 = 0.375
\]

2. For \( h = 0.1 \):

\[
f(1 + 0.1) = f(1.1) = \frac{(1.1)^2}{8}
\]

\[
\text{Average Rate of Change} = \frac{\frac{(1.1)^2}{8} - \frac{1}{8}}{0.1}
\]

3. For \( h = 0.01 \):

\[
f(1 + 0.01) = f(1.01) = \frac{(1.01)^2}{8}
\]

\[
\text{Average Rate of Change} = \frac{\frac{(1.01)^2}{8} - \frac{1}{8}}{0.01}
\]

4. For \( h = 0.001 \):

\[
f(1 + 0.001) = f(1.001) = \frac{(1.001)^2}{8}
\]

\[
\text{Average Rate of Change} = \frac{\frac{(1.001)^2}{8} - \frac{1}{8}}{0.001}
\]

5. For \( h = 0.0001 \):

\[
f(1 + 0.0001) = f(1.0001) = \frac{(1.0001)^2}{8}
\]

\[
\text{Average Rate of Change} = \frac{\frac{(1.0001)^2}{8} - \frac{1}{8}}{0.0001}
\]

Now, I will calculate these values numerically for each value of \( h \).Here are the average rates of change of the function \( f(x) = \frac{x^2}{8} \) over the intervals \([1, 1 + h]\) for each given \( h \), rounded to seven decimal places:

1. For \( h = 1 \): \( 0.3750000 \)
2. For \( h = 0.1 \): \( 0.2625000 \)
3. For \( h = 0.01 \): \( 0.2512500 \)
4. For \( h = 0.001 \): \( 0.2501250 \)
5. For \( h = 0.0001 \): \( 0.2500125 \)

These values show how the average rate of change approaches the instantaneous rate of change as \( h \) gets smaller.

Do you want more details or have any questions?

Here are some related questions you might find interesting:

1. How does the average rate of change relate to the derivative of the function?
2. What is the instantaneous rate of change of the function at \( a = 1 \)?
3. How would these calculations change if \( a \) was a different value, say 2?
4. Can you explain why the average rate of change approaches the derivative as \( h \) approaches 0?
5. How can you visualize these rates of change graphically on a plot of the function?

**Tip:** As \( h \) gets smaller, the average rate of change over the interval \([a, a+h]\) gets closer to the derivative of the function at \( x = a \).

Calculate the average rate of change of the given function f(x) = (x^2)/8 over the intervals [a, a + h] where h = 1, 0.1, 0.01, 0.001, and 0.0001. Round your answers to seven decimal places.

Learn how to calculate the average rate of change of the function f(x) = (x^2)/8 over different intervals. The problem involves calculating the rate of change for intervals where h = 1, 0.1, 0.01, 0.001, and 0.0001. Explore step-by-step solutions and understand how the rate approaches the instantaneous rate as h approaches zero. Suitable for high school students.

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