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Imagine drawing a line on a piece of paper without lifting your pen; this is similar to what mathematicians call a continuous function. In more technical terms, a continuous function is one that is unbroken and has no gaps or sudden jumps for all points within its domain. This means you can trace the graph of the function from one point to any other without interruption.

For example, the function given by the equation \( f(x) = |x| \) is a classic illustration of a continuous function over the interval [-5, 5]. No matter what x-value you choose within this range, there is a corresponding y-value. However, as seen in the STEP BY STEP SOLUTION, a function can be continuous and still not fulfill the Mean Value Theorem. This is because the Mean Value Theorem has an additional requirement - differentiability, which is not satisfied by the function \( f(x) = |x| \) at x = 0 due to the sharp vertex known as a cusp.

On the other side, we have non-continuous functions, which do have interruptions like breaks, holes, or jumps in their graphs. Think of it as the function taking a sudden leap or simply vanishing at certain points within its domain. These functions are not smooth and uninterrupted like their continuous counterparts.

For instance, consider the modified function from the exercise, \( f(x) = |x|\text{, } x eq -2 \). This function is not defined at \( x = -2 \) and hence has a 'hole' in its graph at that point, making it non-continuous. Such functions immediately violate the criteria for the Mean Value Theorem because they don't offer a value at every point within a certain interval, which is a violation of the theorem's primary condition for continuity.

Differentiability is a concept that gives us an insight into the smoothness of a function's curve. If a function is differentiable, it means that you can take its derivative at any point in its domain. The derivative represents the rate at which the function's value is changing at that point, or in other words, the slope of the tangent line to the function's graph at that point.

A function that is differentiable doesn't have any sharp corners or cusps, and it can't be vertical anywhere in its domain. Returning to our previous example, while \( f(x) = |x| \) is continuous over [-5, 5], it is not differentiable at x = 0 because of the cusp. At this point, you can't draw a single tangent line since the slope changes abruptly. Thus, even if a function is continuous, it must also be differentiable to satisfy the conditions of the Mean Value Theorem.