Chapter 4: Problem 3
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
Expert verified
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
01
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 \(a>0\).
02
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.
03
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 \(f(-a)=f(a)\).
04
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 \(a>0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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 \((-a, a)\).
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 \(x < 0\), \(f'(x) = -1\)
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:
This function can be mathematically described as:
- \(f(x) = x\), if \(x \geq 0\)
- \(f(x) = -x\), if \(x < 0\)