Math Problem Statement
Solution
Let's analyze the problem step-by-step.
Given:
A population changes according to the formula .
The question asks which of the following statements are always true:
- is less than the average rate of change of between and .
- The average rate of change of between and is 5.4.
Step 1: Differentiate
The general derivative of an exponential function is: Thus, for , we have:
Step 2: Check the given statements
-
First statement (i): The given form of the derivative in (i) matches the definition of the derivative (using the difference quotient). Therefore, statement (i) is correct.
-
Second statement (ii):
Let's calculate the average rate of change between and . The average rate of change formula is: Now, .Since is greater than 4.4, statement (ii) is false.
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Third statement (iii): We already calculated that the average rate of change between and is 4.4, not 5.4. Hence, statement (iii) is also false.
Conclusion:
The correct answer is (b) none of them, since both (ii) and (iii) are false.
Would you like further details or have any questions?
Related Questions:
- How do you find the derivative of any exponential function ?
- Why does the average rate of change differ from the instantaneous rate of change in this example?
- How does the natural logarithm play a role in differentiating exponential functions?
- Can the average rate of change ever equal the instantaneous rate of change for an exponential function?
- What is the difference between average and instantaneous rates of change in real-life scenarios?
Tip:
Always remember that the derivative gives the instantaneous rate of change, while the average rate of change compares the function's values over an interval.
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Math Problem Analysis
Mathematical Concepts
Differentiation
Exponential Growth
Rates of Change
Formulas
P(t) = 5.4^t
Derivative of an exponential function: d/dt[a^t] = a^t ln(a)
Average rate of change = [P(1) - P(0)] / [1 - 0]
Theorems
Derivative of exponential functions
Definition of average rate of change
Suitable Grade Level
College level (Calculus I)
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