Question

Explain why Rolle's Theorem cannot be applied to the function \(f(x)=|x|\) on the interval \([-a, a],\) for any \(a>0\)

Short answer

Answer: Rolle's Theorem cannot be applied to the function f(x) = |x| on the interval [-a, a], for any a > 0 because the function is not differentiable on the open interval (-a, a).

Step-by-step solution

Define the given function and interval

The function is \(f(x)=|x|\), and we are trying to apply Rolle's Theorem on the interval \([-a,a]\) for any \(a>0\).

Point out the conditions of Rolle's Theorem

As per Rolle's Theorem, three conditions need to be satisfied: 1. The function f(x) must be continuous on the interval [a, b]. 2. The function f(x) must be differentiable on the open interval (a, b). 3. f(a) = f(b), i.e., the function must have the same value at both endpoints of the interval.

Check if the conditions of Rolle's Theorem are satisfied

1. The given function \(f(x)=|x|\) is continuous for all x. Therefore, f(x) is continuous on the interval \([-a, a]\) for any \(a>0\). 2. The derivative of \(f(x)=|x|\) is \(f'(x)=\begin{cases} 1, & x>0 \\ -1, & x<0 \end{cases}\). Since the derivative does not exist at x = 0, the function is not differentiable on the open interval \((-a, a)\). 3. Since \(f(-a)=|-a|=a\) and \(f(a)=|a|=a\) (for \(a>0\)), it is true that \(f(-a)=f(a)\).

State the conclusion and explain why Rolle's Theorem cannot be applied

Since the function \(f(x)=|x|\) is not differentiable on the open interval \((-a, a)\), it does not meet all the requirements of Rolle's Theorem. Thus, Rolle's Theorem cannot be applied to the function \(f(x)=|x|\) on the interval \([-a, a]\), for any \(a>0\).

Key concepts

continuity

Continuity is a fundamental concept in calculus that deals with the behavior of functions across an interval. A function is considered continuous if it does not have any sudden jumps, breaks, or holes in its graph within a given range. This means you can draw the graph of the function without lifting your pencil off the paper. In mathematical terms, a function \(f(x)\) is continuous on an interval \([a, b]\) if for every value \(x\) in that interval, \( \lim_{{x \to c}} f(x) = f(c) \) where \( c \) is any point within \([a, b]\).

For the absolute value function \(f(x) = |x|\), it is continuous across the interval \([-a, a]\) because it does not have any abrupt changes in direction or undefined points. Despite its sharp corner at \(x = 0\), it is still considered continuous because the function does not "break" or become undefined at any point within the specified interval.

differentiability

Differentiability is the quality of a function that determines whether it has a defined derivative at every point within an interval. Differentiability is directly related to the smoothness of a function's graph. If a graph has a sharp corner, cusp, or a vertical tangent line, the function is not differentiable at that point. The derivative of a function at any point \(x\) is the limit of the difference quotient, \( \lim_{{h \to 0}} \frac{f(x + h) - f(x)}{h} \).

The absolute value function \(f(x) = |x|\) presents an interesting case in terms of differentiability. While continuous, it is not differentiable at \(x = 0\) because there is a sharp point at this location. The derivative \(f'(x)\) exists for \(x eq 0\) as it forms two distinct linear parts: a positive slope for \(x > 0\) and a negative slope for \(x < 0\). The transition at \(x = 0\) between these two slopes creates a sharp corner, which is why the function is not differentiable across the whole interval \((-a, a)\).

• For \(x > 0\), \(f'(x) = 1\)
• For \(x < 0\), \(f'(x) = -1\)

However, at \(x = 0\), \(f'(x)\) does not exist, thus breaking the requirement of differentiability.

absolute value function

The absolute value function, represented as \(f(x) = |x|\), is a widely used piecewise function in mathematics. It outputs the magnitude of a real number \(x\), effectively stripping away its sign and leaving only its size. When plotted on a graph, the absolute value function has a V-shaped graph, reflecting all negative input values to their positive counterparts at the same distance from zero.

This function can be mathematically described as:

• \(f(x) = x\), if \(x \geq 0\)
• \(f(x) = -x\), if \(x < 0\)

The sharp point or vertex at \(x = 0\) is a hallmark of the absolute value function and is the primary reason why it is not differentiable at this point. Despite this, it remains continuous everywhere, which showcases an important distinction between continuity and differentiability. While every differentiable function is continuous, not every continuous function is differentiable, as demonstrated by the absolute value function.