91Ó°ÊÓ

Continuous functions are the backbone of many mathematical theorems, including the Mean Value Theorem. Essentially, a function is continuous on an interval if you can draw it on a graph without lifting your pencil. For a function to be continuous, it must satisfy three conditions:

• The function must be defined at every point in the interval.
• Limits from both sides of a point must approach the function's value at that point.
• The function's value at any point should equal the limit as the point is approached from either side.

In the given exercise with the function \( f(x) = \sqrt{x-1} \), the interval \([1,3]\) is critical as the function is continuous there. Since \( x \geq 1 \) ensures it doesn't fall below the range of a square root function, there's a smooth transition of values from \( x = 1 \) to \( x = 3 \). Without continuity, applying the Mean Value Theorem wouldn't be possible.

Differentiability is about a function having a derivative at every point within a given interval, which implies smoothness with no sharp bends or breaks. If a function is differentiable on an open interval \((a, b)\), it is smooth enough to have a tangent line at every point.To be differentiable, a function must first be continuous. Moreover, the derivative, representing the function's rate of change, must exist at every point on this interval.
In the original exercise, \( f(x) = \sqrt{x-1} \) behaves well enough on \( (1,3) \) due to being differentiable when \( x > 1 \). By checking differentiability with the derivative \( f^{\prime}(x) = \frac{1}{2\sqrt{x-1}} \), you see it produces valid results within this open interval. If the function wasn't differentiable, the application of Mean Value Theorem would fail.

The derivative of a function provides a measure of how the function's value changes as its input changes. It is especially useful in determining the slope of the tangent line to the function at any point. Calculating derivatives involves rules, such as the power rule and chain rule.
In the given problem, the derivative of \( f(x) = \sqrt{x-1} \) is derived using the power rule:\[ f^{\prime}(x) = \frac{1}{2}(x-1)^{-1/2} = \frac{1}{2\sqrt{x-1}}. \]This derivative tells us how steep the curve of \( f \) is at any point within the interval. Since the Mean Value Theorem requires this derivative at a specific point \( c \), finding \( f^{\prime}(c) \) is essential part of solving the problem. It links the average rate of overall change to an instantaneous rate at the specific point found in the theorem.

Intervals indicate the range for which a mathematical concept like continuity or differentiability applies. They can be open or closed, denoted by parentheses \((a, b)\) or brackets \([a, b]\) respectively.

In our exercise, we look at two types of intervals:

• The closed interval \([1,3]\) where continuity is checked. This includes endpoints 1 and 3, making sure the function is continuous at every point including these boundaries.
• The open interval \((1,3)\) where differentiability is verified. This excludes the endpoints, focusing on the interval between them to ensure smooth function behavior and derivative existence.

These intervals are crucial as they define the region where the Mean Value Theorem can be applied successfully. Understanding the correct type of interval ensures that necessary conditions like continuity and differentiability are met appropriately.