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An average rate of change measures how a quantity changes on average over a certain interval. It's like calculating the slope between two points on a graph. For a function, consider two points, \(a\) and \(b\), and their corresponding function values \(f(a)\) and \(f(b)\). The formula for the average rate of change is given by:

• \[ \frac{f(b) - f(a)}{b - a} \]

The average rate of change provides a straightforward measure of change over a specific interval but does not tell us about the behavior of the function at any single point within that interval. Thus, it's like getting an overall picture between two points without knowing the details in between.For example, in the function in the exercise, the average rate of change between \(x = 1\) and \(x = 4\) was calculated to be \[ \frac{1}{4} \]. This means that on average, for each unit increase in \(x\), the function \(f(x)\) changes by \[ \frac{1}{4} \].

The derivative of a function tells us the rate at which the function's output value changes as its input value changes. In essence, it's a tool for finding instantaneous rates of change—the slope of the tangent line to the curve at any given point. To find the derivative of a function like \(f(x) = (x-3)^{-2}\), we use the power rule:

• The power rule states that if \(f(x) = x^n\), then the derivative \(f'(x) = nx^{n-1}\).

For the function \(f(x) = (x-3)^{-2}\), applying the power rule gives the derivative:

• \[ f'(x) = \frac{-2}{(x-3)^3} \]

This indicates how the function's behavior changes instantaneously at any point \(x\). A negative derivative, like in \(f'(x)\), signifies that the function \(f(x)\) is decreasing as \(x\) increases around that point.

Continuity and differentiability are crucial conditions for applying the Mean Value Theorem. A function is continuous over an interval if there are no breaks, jumps, or holes in the graph within that interval. Differentiability, on the other hand, means that the function has a defined derivative everywhere in the interval.In the context of the Mean Value Theorem, which requires both continuity on \[a, b\] and differentiability on \( (a, b)\), these two conditions ensure that the function behaves predictably without any abrupt changes.The exercise's function \(f(x) = (x-3)^{-2}\) is not continuous on the specified interval \[1, 4\] due to a vertical asymptote at \(x = 3\). This makes the function neither continuous nor differentiable over \[1, 4\], invalidating the use of the Mean Value Theorem in this context. Without fulfilling these conditions, we cannot guarantee the existence of a point where the instantaneous rate of change equals the average rate of change.

A vertical asymptote in a function occurs at a particular point where the function heads towards infinity as the input approaches a specific value. In simple terms, it is a vertical line that the graph of the function approaches but never actually touches.For \(f(x) = (x-3)^{-2}\), there is a vertical asymptote at \(x = 3\). As \(x\) inches closer to 3 from either the left or the right, the values of \(f(x)\) shoot up towards infinity. This behavior disrupts continuity, as the function isn’t defined at \(x = 3\). Vertical asymptotes are significant because:

• They indicate points of discontinuity in a function.
• They prevent the application of certain mathematical theorems, like the Mean Value Theorem, across an interval containing the asymptote.

Therefore, because of this vertical asymptote, the function \(f(x) = (x-3)^{-2}\) doesn't satisfy the conditions necessary to apply the Mean Value Theorem on the interval \[1, 4\].