Question

Determine whether Rolle's Theorem can be applied to \(f\) on the closed interval \([a, b] .\) If Rolle's Theorem can be applied, find all values of \(c\) in the open interval \((a, b)\) such that \(f^{\prime}(c)=0\). $$ f(x)=3-|x-3|,[0,6] $$

Short answer

Rolle's Theorem cannot be applied to the function on the interval \([0,6]\), as the function is not differentiable on the entire interval \((0,6)\), specifically at \(x = 3\). Therefore, there's no value of \(c\) in this interval where \(f^{\prime}(c) = 0\).

Step-by-step solution

Evaluate the Function at the Endpoints

The first thing to do is to evaluate \(f(x)\) at \(x = a\) and \(x = b\). Here, \(f(x) = 3 - |x - 3|\) and the interval given is \([0, 6]\), so \(a = 0\) and \(b = 6\). Evaluating, we find that \(f(0) = f(6) = 3\). Thus, the first condition of Rolle's Theorem is satisfied.

Check for Continuity and Differentiability

The function is continuous on its entire domain, which includes \([0,6]\). However, we need to determine if it's differentiable on the open interval \((0,6)\). The derivative of the function is dependent on whether \(x < 3\) or \(x > 3\) because of the absolute value. The derivative is \(f^{\prime}(x) = 1\) for \(x < 3\) and \(f^{\prime}(x) = -1\) for \(x > 3\). The function is not differentiable at \(x = 3\) because the derivative at this point is not defined, hence it does not satisfy all conditions for Rolle's theorem.

Determine if Rolle's Theorem Applies

Since the function is not differentiable on the entire open interval \((0,6)\), specifically at \(x = 3\), it fails to meet all the conditions of Rolle's Theorem. Therefore, Rolle's Theorem cannot be applied and there is no value of \(c\) in the interval \((0, 6)\) where \(f^{\prime}(c) = 0\).

Key concepts

Differentiability

In calculus, differentiability is the property that allows us to determine the rate at which a function's value changes at a given point. For a function to be differentiable at a point, it must have a unique tangent line at that point, implying that the function's behavior is smooth and continuous there.

To check differentiability, one usually looks for the existence of a derivative, which represents the slope of the tangent. Mathematically, if the limit
\[\begin{equation}\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\end{equation}\]exists and is finite, the function is differentiable at point 'x'. However, a function with a sharp turn or corner, which might yield two different slopes depending on the direction from which 'x' is approached, is considered not differentiable at that point.

In the case of the function \(f(x) = 3 - |x - 3|\), the problem arises at \(x = 3\), where there is a 'kink' or sharp turn. This is due to the nature of the absolute value function, which has a corner at the point where its argument is zero. Subsequently, differentiability fails at \(x = 3\), and as Rolle's Theorem requires the function to be differentiable on the entire open interval, it cannot be applied in this scenario.

Absolute Value

The absolute value of a real number is a non-negative value that represents its distance from zero along the number line, regardless of its direction. Represented by two vertical lines surrounding a number or expression, \(|x|\), it turns negative inputs into positive outputs, while leaving non-negative inputs unchanged.

Graphically, the absolute value function creates a 'V' shape, with a sharp corner at the point where the input is zero. For example, the function \(y = |x|\) has such a corner at the origin, \((0, 0)\). Due to this characteristic shape, the absolute value function causes issues with differentiability at its corner point, because it changes direction abruptly, which prevents the existence of a unique tangent line.

Differentiability of Absolute Value Functions

Mathematically, differentiating an absolute value function requires considering the sign of the input. For \(f(x) = 3 - |x - 3|\), when \(x < 3\), the function is equivalent to \(3 - (3 - x)\), which simplifies to \(x\), while for \(x > 3\), it is equivalent to \(3 - (x - 3)\), which simplifies to \(6 - x\). Therefore, its derivative changes abruptly at \(x = 3\), where the 'V' shape's corner is located, leading to the function being non-differentiable at that point.

Continuity

Continuity in mathematics is about the smooth progression of a function. For a function to be continuous at a point, the following three conditions must be met: The function must be defined at that point, the limit of the function as it approaches the point from both directions must exist, and the function’s limit as it approaches the point must equal the function’s actual value at that point.

This means that you can draw the graph of a continuous function without lifting your pencil from the paper. Specifically, for Rolle's Theorem to apply, a function needs to be continuous on the closed interval \([a, b]\).

The function \(f(x)=3-|x-3|\) meets the criteria for continuity on the given interval \([0, 6]\). Even at \(x = 3\), where the absolute value alters direction, the function's value smoothly transitions from one linear piece to the other, without any breaks or jumps. This satisfies the definition of continuity for the interval, but as discussed earlier, continuity alone is not enough to fulfill the conditions of Rolle's Theorem without differentiability on the open interval \((a, b)\).