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?
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 the function \( g(x) = 25x \), we first need to compute the average rates of change for both functions over an interval \([a, b]\).
The average rate of change of a function \( h(x) \) over the interval \([a, b]\) is given by
\[
\text{Average Rate of Change} = \frac{h(b) - h(a)}{b - a}
\]
### Step 1: Calculate the average rate of change for \( f(x) \)
Applying this to \( f(x) = 5^x \):
\[
\text{Average Rate of Change of } f \text{ over } [a, b] = \frac{5^b - 5^a}{b - a}
\]
### Step 2: Calculate the average rate of change for \( g(x) \)
Now for \( g(x) = 25x \):
\[
\text{Average Rate of Change of } g \text{ over } [a, b] = \frac{25b - 25a}{b - a} = \frac{25(b - a)}{b - a} = 25
\]
### Step 3: Set up the inequality
We need to find intervals where:
\[
\frac{5^b - 5^a}{b - a} > 25
\]
### Step 4: Simplify the inequality
We can rewrite this as:
\[
5^b - 5^a > 25(b - a)
\]
### Step 5: Analyze the original functions
To solve \( 5^b - 5^a > 25(b - a) \), we note that as \( a \) and \( b \) approach each other, the expression can be analyzed using calculus. Let's consider the derivative of \( f(x) = 5^x \):
\[
f'(x) = 5^x \ln(5)
\]
For \( g(x) = 25x \):
\[
g'(x) = 25
\]
We are looking for when the instantaneous rate of change \( f'(x) \) is greater than \( g'(x) \):
\[
5^x \ln(5) > 25
\]
### Step 6: Solve the inequality
Dividing both sides by \( \ln(5) \) (which is positive):
\[
5^x > \frac{25}{\ln(5)}
\]
Taking the logarithm base \( 5 \) of both sides:
\[
x > \log_5\left(\frac{25}{\ln(5)}\right)
\]
Since \( 25 = 5^2 \):
\[
x > 2 - \log_5(\ln(5))
\]
### Conclusion
The average rate of change of \( f(x) = 5^x \) is greater than that of \( g(x) = 25x \) for the intervals \( x > 2 - \log_5(\ln(5)) \). Thus, the final result indicates the intervals of \( x \):
\[
x > 2 - \log_5(\ln(5))
\]
The average rate of change of a function \( h(x) \) over the interval \([a, b]\) is given by
\[
\text{Average Rate of Change} = \frac{h(b) - h(a)}{b - a}
\]
### Step 1: Calculate the average rate of change for \( f(x) \)
Applying this to \( f(x) = 5^x \):
\[
\text{Average Rate of Change of } f \text{ over } [a, b] = \frac{5^b - 5^a}{b - a}
\]
### Step 2: Calculate the average rate of change for \( g(x) \)
Now for \( g(x) = 25x \):
\[
\text{Average Rate of Change of } g \text{ over } [a, b] = \frac{25b - 25a}{b - a} = \frac{25(b - a)}{b - a} = 25
\]
### Step 3: Set up the inequality
We need to find intervals where:
\[
\frac{5^b - 5^a}{b - a} > 25
\]
### Step 4: Simplify the inequality
We can rewrite this as:
\[
5^b - 5^a > 25(b - a)
\]
### Step 5: Analyze the original functions
To solve \( 5^b - 5^a > 25(b - a) \), we note that as \( a \) and \( b \) approach each other, the expression can be analyzed using calculus. Let's consider the derivative of \( f(x) = 5^x \):
\[
f'(x) = 5^x \ln(5)
\]
For \( g(x) = 25x \):
\[
g'(x) = 25
\]
We are looking for when the instantaneous rate of change \( f'(x) \) is greater than \( g'(x) \):
\[
5^x \ln(5) > 25
\]
### Step 6: Solve the inequality
Dividing both sides by \( \ln(5) \) (which is positive):
\[
5^x > \frac{25}{\ln(5)}
\]
Taking the logarithm base \( 5 \) of both sides:
\[
x > \log_5\left(\frac{25}{\ln(5)}\right)
\]
Since \( 25 = 5^2 \):
\[
x > 2 - \log_5(\ln(5))
\]
### Conclusion
The average rate of change of \( f(x) = 5^x \) is greater than that of \( g(x) = 25x \) for the intervals \( x > 2 - \log_5(\ln(5)) \). Thus, the final result indicates the intervals of \( x \):
\[
x > 2 - \log_5(\ln(5))
\]