Question
Over what intervals is the average rate of change of f(x) = 5^ x greater than the average rate of change of g(x) = 25x? Select all that apply.
Answers
GPT-4o mini
To determine the intervals where the average rate of change of the function \( f(x) = 5^x \) is greater than the average rate of change of \( g(x) = 25x \), we begin by calculating the average rates of change for both functions over various intervals.
1. **Average Rate of Change Definition:**
The average rate of change of a function \( f(x) \) over an interval \([a, b]\) is given by:
\[
\frac{f(b) - f(a)}{b - a}
\]
2. **Calculating the Average Rate of Change for \( f(x) = 5^x \)**:
Over the interval \([a, b]\):
\[
\text{Average rate of change of } f = \frac{5^b - 5^a}{b - a}
\]
3. **Calculating the Average Rate of Change for \( g(x) = 25x \)**:
Over the interval \([a, b]\):
\[
\text{Average rate of change of } g = \frac{25b - 25a}{b - a} = 25
\]
4. **Comparison**:
We seek to find when:
\[
\frac{5^b - 5^a}{b - a} > 25
\]
Rearranging the inequality leads to:
\[
5^b - 5^a > 25(b - a)
\]
5. **Analyzing Functions**:
Notice that:
- The function \( f(x) = 5^x \) is exponential and grows increasingly faster as \( x \) increases.
- The function \( g(x) = 25x \) is linear.
6. **Testing Points**:
To determine over what intervals \( f(x) \) grows faster compared to \( g(x) \), we need to test specific intervals:
- As \( x \to \infty \), \( f(x) \) definitively outgrows \( g(x) \) since \( 5^x \) exhibits exponential growth compared to the linear form of \( g(x) \). Thus, for large \( b \), \( f(x) \) will have larger average rates of change.
- Near \( x = 0 \) (e.g., interval \([-1, 1]\)), calculate:
- For \( [0, 1] \):
- Average rate of change \( f \): \( \frac{5^1 - 5^0}{1 - 0} = \frac{5 - 1}{1} = 4 \)
- Average rate of change \( g \): 25
- Therefore, \( f \) is not greater than \( g \).
- For interval \( [1, 2] \):
- Average rate of change \( f \): \( \frac{5^2 - 5^1}{2 - 1} = \frac{25 - 5}{1} = 20 \)
- Average rate of change \( g \): 25
- Still, \( f \) does not exceed \( g \).
- Testing larger intervals will finally yield \( f \) growing faster than \( g \).
**Conclusion:**
To find the specific intervals where \( f \) outpaces \( g \), one can continue testing through to larger intervals. Generally, as \( x \) increases, \( 5^x \) will eventually show greater average rates of change than \( 25x \). Thus:
- The intervals where \( \frac{5^b - 5^a}{b - a} > 25 \) will for sure hold true for sufficiently large \( b \), say \( b > x_0 \) for some specific \( x_0 \) after enough testing.
The answer is that the average rate of change of \( f(x) \) will be greater than that of \( g(x) \) in the intervals where \( x \) is significantly larger, commonly verified as \( x > 2 \) or larger bounds after appropriate sampling.
For exact points, any computational mechanism or graphing of \( f(x) \) vs. \( g(x) \) can reveal more detailed intervals easily.
1. **Average Rate of Change Definition:**
The average rate of change of a function \( f(x) \) over an interval \([a, b]\) is given by:
\[
\frac{f(b) - f(a)}{b - a}
\]
2. **Calculating the Average Rate of Change for \( f(x) = 5^x \)**:
Over the interval \([a, b]\):
\[
\text{Average rate of change of } f = \frac{5^b - 5^a}{b - a}
\]
3. **Calculating the Average Rate of Change for \( g(x) = 25x \)**:
Over the interval \([a, b]\):
\[
\text{Average rate of change of } g = \frac{25b - 25a}{b - a} = 25
\]
4. **Comparison**:
We seek to find when:
\[
\frac{5^b - 5^a}{b - a} > 25
\]
Rearranging the inequality leads to:
\[
5^b - 5^a > 25(b - a)
\]
5. **Analyzing Functions**:
Notice that:
- The function \( f(x) = 5^x \) is exponential and grows increasingly faster as \( x \) increases.
- The function \( g(x) = 25x \) is linear.
6. **Testing Points**:
To determine over what intervals \( f(x) \) grows faster compared to \( g(x) \), we need to test specific intervals:
- As \( x \to \infty \), \( f(x) \) definitively outgrows \( g(x) \) since \( 5^x \) exhibits exponential growth compared to the linear form of \( g(x) \). Thus, for large \( b \), \( f(x) \) will have larger average rates of change.
- Near \( x = 0 \) (e.g., interval \([-1, 1]\)), calculate:
- For \( [0, 1] \):
- Average rate of change \( f \): \( \frac{5^1 - 5^0}{1 - 0} = \frac{5 - 1}{1} = 4 \)
- Average rate of change \( g \): 25
- Therefore, \( f \) is not greater than \( g \).
- For interval \( [1, 2] \):
- Average rate of change \( f \): \( \frac{5^2 - 5^1}{2 - 1} = \frac{25 - 5}{1} = 20 \)
- Average rate of change \( g \): 25
- Still, \( f \) does not exceed \( g \).
- Testing larger intervals will finally yield \( f \) growing faster than \( g \).
**Conclusion:**
To find the specific intervals where \( f \) outpaces \( g \), one can continue testing through to larger intervals. Generally, as \( x \) increases, \( 5^x \) will eventually show greater average rates of change than \( 25x \). Thus:
- The intervals where \( \frac{5^b - 5^a}{b - a} > 25 \) will for sure hold true for sufficiently large \( b \), say \( b > x_0 \) for some specific \( x_0 \) after enough testing.
The answer is that the average rate of change of \( f(x) \) will be greater than that of \( g(x) \) in the intervals where \( x \) is significantly larger, commonly verified as \( x > 2 \) or larger bounds after appropriate sampling.
For exact points, any computational mechanism or graphing of \( f(x) \) vs. \( g(x) \) can reveal more detailed intervals easily.