Question

On a graph points are (-6,13) and (5,0) calculate the an average rate of change of the exponential function over the given interval. Estimate the average rate of change go f(x)=(x-0.1)^2-24.01 over the same interval how do they compare

Answer 1

To calculate the average rate of change of the exponential function over the interval defined by the points \((-6, 13)\) and \( (5, 0)\), we follow these steps:

### Step 1: Average Rate of Change of the Exponential Function

The average rate of change of a function \( f(x) \) over an interval \([a, b]\) is given by the formula:

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

For the given points, \( a = -6 \) and \( b = 5 \).

### Step 2: Values of the Function

Assuming we need to calculate average rate of change for an exponential function \( f(x) = e^x \), we'll compute the values:

- Calculate \( f(-6) \):
\[
f(-6) = e^{-6} \approx 0.002478752
\]

- Calculate \( f(5) \):
\[
f(5) = e^5 \approx 148.41316
\]

### Step 3: Average Rate of Change Calculation

Now substitute these values into the average rate of change formula:

\[
\text{Average Rate of Change} = \frac{f(5) - f(-6)}{5 - (-6)} = \frac{148.41316 - 0.002478752}{5 + 6}
\]

\[
= \frac{148.41068}{11} \approx 13.5491
\]

### Step 4: Estimate the Average Rate of Change of \( f(x) = (x - 0.1)^2 - 24.01 \)

Next, we find the average rate of change of the given quadratic function over the same interval.

First, compute \( f(-6) \) and \( f(5) \):

- Calculation of \( f(-6) \):
\[
f(-6) = (-6 - 0.1)^2 - 24.01 = (-6.1)^2 - 24.01 = 37.21 - 24.01 = 13.20
\]

- Calculation of \( f(5) \):
\[
f(5) = (5 - 0.1)^2 - 24.01 = (4.9)^2 - 24.01 = 24.01 - 24.01 = 0
\]

### Step 5: Average Rate of Change Calculation for the Quadratic Function

Now substitute these values into the average rate of change formula:

\[
\text{Average Rate of Change} = \frac{f(5) - f(-6)}{5 - (-6)} = \frac{0 - 13.20}{5 + 6}
\]

\[
= \frac{-13.20}{11} \approx -1.20
\]

### Step 6: Comparison

- Average Rate of Change of the assumed exponential function: \(\approx 13.5491\)
- Average Rate of Change of the quadratic function: \(\approx -1.20\)

### Conclusion

The average rate of change for the exponential function \( e^x \) over the interval is significantly greater than the average rate of change of the quadratic function \( (x - 0.1)^2 - 24.01 \). This reflects the rapid growth of the exponential function compared to the behavior of the quadratic function in the given interval.