91Ó°ÊÓ

A continuous function is one that has no breaks, holes, or jumps in its graph. Mathematically, a function \(f(x)\) is continuous on an interval \[a, b\] if for every point \(x\) in \[a, b\], the limit of \(f(x)\) as \(x\) approaches any point \(c\) within this interval equals the value of the function at that point. Here’s a simpler way to understand this concept:

• If you can draw the graph of the function without lifting your pen, it's continuous.
• An example of a continuous function is our given function \(f(x) = x^3 - 9x + 1\) on the interval \[-3, 4\].

Continuous functions are vital for the Mean Value Theorem, as the theorem guarantees the existence of a point where the slope of the tangent is equal to the slope of the secant line within the interval.

A differentiable function has a derivative at every point within its domain. In simpler terms, it means the function has a defined slope at each point. For \(f(x)\) to be differentiable on \[a, b\], it must also be continuous on that interval. Here’s what this means:

• The function should not have any sharp corners.
• No vertical tangent lines should exist.

If a function is differentiable, it ensures smoothness, meaning you can calculate the exact rate of change at any point. For our given function \(f(x) = x^3 - 9x + 1\), we computed the derivative as \(f'(x) = 3x^2 - 9\). This is a polynomial, which is differentiable everywhere. Therefore, \(f(x)\) meets the second requirement of the Mean Value Theorem.

Understanding tangent and secant lines is key to visualizing the Mean Value Theorem.
A secant line intersects a curve at two points, providing an average rate of change between those points. For our function, the secant line connecting the points at \(a = -3\) and \(b = 4\) has a slope of 4. The equation for this line can be found as follows:
\[ m_{secant} = \frac{f(b) - f(a)}{b - a} = 4 \] On the other hand, a tangent line touches the curve at exactly one point, representing the instantaneous rate of change at that point. According to the Mean Value Theorem, at some point \(c\) within \(-3 < c < 4\), the slope of the tangent line equals that of the secant line:
\( f'(c) = 4 \). In this problem, solving \(3c^2 - 9 = 4\) gives us \( c = \frac{\sqrt{39}}{3} \). Thus, the tangent line at \(c\) confirms the theorem, providing a precise connection between overall and instantaneous change.