Question

lionel computed the average rate of change in the depth of a pool over a two - week interval to be zero. which statement must be true?
the pool must have been empty for the entire interval.
the pool must have been the same depth at the start of the interval as it was at the end of the interval.
the pool must have been deeper at the end of the interval than it was at the start of the interval.
the pool must have been more shallow at the end of the interval than it was at the start of the interval.

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function \(y = f(x)\) over an interval \([a,b]\) is given by \(\frac{f(b)-f(a)}{b - a}\). If the average rate of change is zero, then \(\frac{f(b)-f(a)}{b - a}=0\). Since \(b - a
eq0\) (as it is a two - week interval, \(b>a\)), we have \(f(b)-f(a)=0\), which means \(f(a)=f(b)\).

Step2: Interpret the result in the context of the pool's depth

Let \(f(x)\) be the depth of the pool at time \(x\). If the average rate of change of the depth of the pool over a two - week interval is zero, then \(f(\text{end of interval})-f(\text{start of interval}) = 0\), or \(f(\text{end of interval})=f(\text{start of interval})\). This means the pool must have been the same depth at the start of the interval as it was at the end of the interval.

Answer:

The pool must have been the same depth at the start of the interval as it was at the end of the interval.