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The problem involves determining which of the provided statements about the population function \( P(t) = 5.4^t \) are always true. Let’s break it down step by step.

### Given Function:
- \( P(t) = 5.4^t \)
- The statements are related to the **rate of change** of this function, which means we’ll have to analyze its derivative and average rates of change.

#### (i) \( P'(t) = \lim_{h \to 0} \frac{5.4^h(5.4^t - 1)}{h} \)

This is the derivative of the function \( P(t) = 5.4^t \). To simplify, we note that:
\[
P'(t) = \frac{d}{dt}(5.4^t) = 5.4^t \ln(5.4)
\]
This follows the standard derivative rule for exponential functions. So, the derivative of \( P(t) \) at any point \( t \) is \( P'(t) = 5.4^t \ln(5.4) \).

#### (ii) \( P'(1) \) is less than the average rate of change of \( P \) between 0 and 1.

The derivative \( P'(1) = 5.4^1 \ln(5.4) \). Using approximate values:
\[
P'(1) = 5.4 \times \ln(5.4) \approx 5.4 \times 1.686 = 9.1044
\]



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 the population change function P(t) = 5.4^t, and analyze its derivative and average rate of change. This problem includes detailed steps for derivative calculation and compares the instantaneous rate of change at t = 1 with the average rate between t = 0 and t = 1. Suitable for college-level calculus students.

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