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Let's analyze the problem step-by-step.

### Given: 
A population changes according to the formula \( P(t) = 5.4^t \).

The question asks which of the following statements are **always true**:

1. \( P'(t) = \lim\limits_{h \to 0} \frac{5.4^h (5.4^t - 1)}{h} \)
2. \( P'(1) \) is less than the average rate of change of \( P \) between \( t = 0 \) and \( t = 1 \).
3. The average rate of change of \( P \) between \( t = 0 \) and \( t = 1 \) is 5.4.

#### Step 1: Differentiate \( P(t) = 5.4^t \)
The general derivative of an exponential function \( a^t \) is:
\[
\frac{d}{dt}[a^t] = a^t \ln(a)
\]
Thus, for \( P(t) = 5.4^t \), we have:
\[
P'(t) = 5.4^t \ln(5.4)
\]

#### Step 2: Check the given statements

1. **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.

2. **Second statement (ii):**  
   Let's calculate the average rate of change between \( t = 0 \) and \( t = 1 \). The average rate of change formula is:
   \[
   \text{Average rate of change} = \frac{P(1) - P(0)}{1 - 0}
   \]
   \[
   P(1) = 5.4^1 = 5.4, \quad P(0) = 5.4^0 = 1
   \]
   \[
   \text{Average rate of change} = \frac{5.4 - 1}{1} = 4.4
   \]
   Now, \( P'(1) = 5.4 \ln(5.4) \approx 5.4 \times 1.686 = 9.1044 \).
   
   Since \( P'(1) \approx 9.1044 \) is greater than 4.4, statement (ii) is **false**.

3. **Third statement (iii):**
   We already calculated that the average rate of change between \( t = 0 \) and \( t = 1 \) 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:
1. How do you find the derivative of any exponential function \( a^t \)?
2. Why does the average rate of change differ from the instantaneous rate of change in this example?
3. How does the natural logarithm play a role in differentiating exponential functions?
4. Can the average rate of change ever equal the instantaneous rate of change for an exponential function?
5. 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.

A population changes according to the formula P(t) = 5.4^t. Which of the following statements are always true? (i) P'(t) = lim(h->0) [5.4^h(5.4^t - 1)] / h, (ii) P'(1) is less than the average rate of change of P between 0 and 1, (iii) The average rate of change between 0 and 1 is 5.4.

Explore how to calculate both the derivative and average rate of change for the exponential function P(t) = 5.4^t. Includes a detailed step-by-step solution to the problem with key mathematical concepts such as differentiation and exponential growth.

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