91Ó°ÊÓ

In calculus, proofs play a crucial role in verifying the validity of mathematical statements. The Mean-Value Theorem (MVT) is a fundamental result in calculus, providing a formal foundation to many key ideas. It essentially connects the differentiation of a function with the behavior of the function over an interval.
Understanding and applying the MVT requires establishing two conditions:

• The function must be continuous over a closed interval \[ a, b \].
• The function must be differentiable over the open interval \( (a, b) \).

By meeting these conditions, the MVT assures that there is some point \( c \) within \( (a, b) \) where the instantaneous rate of change (the derivative) is equivalent to the average rate of change over the interval. With an understanding of calculus proofs, you can confidently apply such theorems and verify important properties of functions.

A differentiable function is one whose derivative exists at each point within a given interval. This is a prerequisite for applying the Mean-Value Theorem.
A common method of differentiation involves using the rules of differentiation to calculate the derivative of a given function.
In the context of the provided exercise, \( f(x) = \sqrt{x} \) is indeed differentiable over the interval \( (3, 4) \).

• The derivative, obtained using the power rule, is given by \( f'(x) = \frac{1}{2\sqrt{x}} \).
• Having this derivative allows us to explore how rapidly \( f(x) = \sqrt{x} \) changes at any point within \( (3, 4) \).

Differentiability is essential because it ensures that we can seamlessly apply MVT, supporting the calculation of necessary points that demonstrate the theorem.

Continuity of a function across an interval implies that the function does not have any abrupt breaks, jumps, or holes within that span. In simple terms, a function is continuous over an interval if you can draw it without lifting your pencil.
The function \( f(x) = \sqrt{x} \) is indeed continuous on the closed interval \[ 3, 4 \]. This is apparent because the square root function is continuous for all \( x \) values within \[ 0, \infty \].

• To apply the MVT, it's vital first to establish that the function is continuous over the specified interval \[ a, b \].
• Given that \( f(x) = \sqrt{x} \) is continuous, this allows us to investigate further using differentiation and the MVT, as no unexpected behaviors occur within \[ 3, 4 \].

Thus, continuity not only ensures smoothness over the interval but also verifies the necessary conditions for the Mean-Value Theorem to be applicable.