Problem 169 - Calculus Volume 1

Question

Show there is no \(c\) such that \(f(1)-f(-1)=f^{\prime}(c)(2) .\) Explain why the Mean Value Theorem does not apply over the interval \([-1,1]\) . $$ f(x)=\sqrt{|x|} $$

Step-by-Step Solution

Answer

The Mean Value Theorem does not apply because \(f(x)=\sqrt{|x|}\) is not differentiable at \(x=0\).

Step 1: Understand the Mean Value Theorem

The Mean Value Theorem states that if a function \(f\) is continuous over the closed interval \([a, b]\), and differentiable over the open interval \((a, b)\), then there exists at least one \(c\) in \((a, b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\).

Step 2: Verify Continuity on the Interval

The function \(f(x) = \sqrt{|x|}\) is continuous on the interval \([-1, 1]\) because absolute value and square root functions are continuous everywhere over their domain.

Step 3: Determine Differentiability Conditions

To apply the Mean Value Theorem, \(f\) must be differentiable on the open interval \((-1, 1)\). The function \(f(x) = \sqrt{|x|}\) is differentiable for \(x > 0\) and \(x < 0\), but at \(x = 0\), the function changes form from \(-x^{0.5}\) to \(x^{0.5}\), causing a cusp.

Step 4: Evaluate Differentiability at Critical Point

At \(x = 0\), the derivative does not exist because the derivative from the left \((-0.5\sqrt{-x}^-1)\) is undefined and different from the derivative form the right \((0.5\sqrt{x}^{-1})\). Hence, there is no uniform derivative at \(x = 0\).

Step 5: Analyzing the Conclusion

Since \(f(x)\) is not differentiable at \(x = 0\), the requirements for the Mean Value Theorem are not satisfied over the interval \([-1, 1]\). Therefore, there is no \(c\) such that \(f(1)-f(-1)=f^{\prime}(c)(2)\).

Key Concepts

Continuity

The concept of continuity is foundational in calculus, as it ensures a function behaves in a predictable manner without any abrupt changes. A function is said to be continuous over an interval if you can draw it on a graph without lifting your pencil from the paper. In the given exercise, we analyze the function \(f(x) = \sqrt{|x|}\), which is continuous over the interval \([-1, 1]\).

Both the square root and absolute value functions are inherently continuous across their domains. Therefore, despite the absolute value introducing a change at \(x = 0\), the function \(f(x)\) remains continuous across the whole interval. This property is essential for the Mean Value Theorem, which requires continuity on a closed interval, like \([-1, 1]\).

• No jumps or holes exist in the function \(f(x)\).
• You can easily graph \(f(x) = \sqrt{|x|}\) without interruptions.

Differentiability

Differentiability is a step further than continuity, requiring a function not only to be smooth but also to have a well-defined slope at every point on an interval. A function that is differentiable at a point must also be continuous there, but the reverse isn't always true.

In our problem, the function \(f(x) = \sqrt{|x|}\) is differentiable over \((a, b) = (-1, 1)\), except at \(x = 0\). This is due to the cusp at \(x = 0\), where the behavior of the function changes sharply. Differentiability fails here because:

• The derivative from the left-hand side, involving \(-x^{0.5}\), doesn't match the right-hand side involving \(x^{0.5}\).
• There isn't a unique tangent line slope at \(x = 0\).

These characteristics of the cusp render the function non-differentiable at \(x = 0\), preventing the Mean Value Theorem from holding.

Interval

An interval in mathematics denotes a range of values that a particular variable can take. When evaluating the Mean Value Theorem, we must carefully assess the nature of the interval involved. In this exercise, we're working with the interval \([-1, 1]\).

For the Mean Value Theorem to be applicable, a function should be continuous over a closed interval like \([-1, 1]\), and differentiable over an open interval like \((-1, 1)\).

• The closed interval \([-1, 1]\) includes all points from \(-1\) to \(1\), while the open interval \((-1, 1)\) excludes the endpoints.
• Problems arise within these intervals if any point like a cusp ensues, which invalidates differentiability.

The inconsistency of differentiability at any point like \(x = 0\) causes the requirements of the Mean Value Theorem not to be fully met.

Cusp

A cusp in the context of functions is a point on the graph where two sections of the curve meet and form a sharp corner. These corners disrupt the smooth nature needed for differentiability but do not necessarily interrupt continuity.

When considering \(f(x) = \sqrt{|x|}\), a cusp is found at \(x = 0\). Here’s why:

• The graph changes from descending to ascending sharply at \(x = 0\).
• Derivative calculations on either side of \(x = 0\) produce different and undefined results.

Because of the cusp at \(x = 0\), the derivative does not exist at this point. Hence, the Mean Value Theorem does not apply over \([-1, 1]\), because differentiability is a necessary condition, and it's violated at the cusp.