Problem 12 - Thomas Calculus in SI Units

Question

Which of the functions satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers. $$f(x)=\left\\{\begin{array}{ll}2 x-3, & 0 \leq x \leq 2 \\\6 x-x^{2}-7, & 2

Step-by-Step Solution

Answer

The function satisfies the Mean Value Theorem on [0, 3].

Step 1: Understanding the Mean Value Theorem

The Mean Value Theorem (MVT) states that if a function \( f(x) \) is continuous on the closed interval \([a, b] \) and differentiable on 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: Check Continuity

Check if the function \( f(x) \) is continuous on the interval [0, 3]. This involves checking if both pieces of the piecewise function meet the requirement for continuity, particularly at the joining point \( x = 2 \). Evaluate the limits of both pieces at \( x = 2 \) to ensure continuity.

Step 3: Verify \( f(x) \) at \( x = 2 \) is Continuous

For continuity, \( \lim_{x \to 2^-} f(x) = f(2) = \lim_{x \to 2^+} f(x) \). Evaluate \( \lim_{x \to 2^-} (2x-3) = 1 \), \( f(2)=1 \), and \( \lim_{x \to 2^+} (6x-x^2-7) = 1 \). Thus, the function is continuous at \( x = 2 \).

Step 4: Check Differentiability

Now, we will check if the function is differentiable on the open interval \((0, 3)\). First, ensure that the derivatives of both pieces exist where they apply. Then, check the point \( x = 2 \) where the pieces meet for differentiability by ensuring the derivative from the left equals the derivative from the right.

Step 5: Evaluate Derivatives

Compute the derivative for the first piece \( f(x)=2x-3 \), which is \( f'(x) = 2 \) for \( 0 < x < 2 \). For the second piece \( f(x)=6x-x^2-7 \), the derivative is \( f'(x) = 6 - 2x \) for \( 2 < x < 3 \).

Step 6: Check Derivative at \( x = 2 \)

Evaluate the left derivative limit, \( \lim_{x \to 2^-} f'(x) = 2 \) and the right derivative limit, \( \lim_{x \to 2^+} f'(x) = 2 \). They are equal, hence the function is differentiable at \( x = 2 \).

Step 7: Conclusion about Mean Value Theorem Applicability

Since \( f(x) \) is continuous on \([0, 3]\) and differentiable on \((0, 3)\), the function satisfies the hypotheses of the Mean Value Theorem on this interval.

Key Concepts

Continuity

Continuity is a fundamental concept in calculus that ensures a smooth transition of a function without any breaks, jumps, or holes at any point in its domain. For a function to be continuous on a closed interval \([a, b]\), it must not only be continuous at every point within that interval but also have the limits from both ends matching the actual function values at points \(a\) and \(b\).
For piecewise functions like the one in this exercise, checking continuity involves ensuring that there are no jumps between different "pieces" of the function, particularly at their joining points. In our exercise, we need to check continuity at the point \(x=2\).
To do this:

• Evaluate \(\lim_{x \to 2^-}(2x-3)\) for values approaching \(x=2\) from the left.
• Ensure this limit equals the actual function value at \(x=2\), which in this case is \(f(2) = 1\).
• Verify \(\lim_{x \to 2^+}(6x-x^2-7)\) for values approaching \(x=2\) from the right.
• All these values should match to confirm the continuity of the entire piecewise function \(f(x)\) over the interval.

Successfully establishing these equalities confirms the function's continuity at this crucial checking point, thus fulfilling one of the core requirements for applying the Mean Value Theorem.

Differentiability

Differentiability is about the ability to take a derivative of a function at given points, providing insights into its behavior through slopes of tangents. A function is differentiable at a point if its derivative exists at that point and the function is continuous in the neighborhood of the point. In the context of piecewise functions, differentiability is often checked at the points where the function pieces meet.
For the function given in this exercise, the differentiability needs to be assessed at the point \(x=2\) (the point of transition between the two function pieces) and across the open interval \((0, 3)\):

• Calculate the derivative of the first piece \(f(x)=2x-3\), yielding \(f'(x) = 2\) for \(x\) within the first interval.
• Find the derivative for the second piece \(f(x)=6x-x^2-7\), resulting in \(f'(x) = 6 - 2x\) for \(x\) within the second interval.
• Verify that the derivative as \(x\) approaches \(2\) from the left (\(\lim_{x \to 2^-}f'(x)\)) equals the derivative from the right (\(\lim_{x \to 2^+}f'(x)\)), both equating to \(2\).

Matching derivative values ensure that the function behaves smoothly at the transition point, confirming differentiability along the open interval \((0, 3)\), which is essential for applying the Mean Value Theorem.

Piecewise Functions

Piecewise functions are made up of "pieces" or segments, each with its own defining equation applicable over different parts of the function's domain. These functions can model situations where behavior changes at specific points, providing flexibility in representing complex real-world phenomena through simpler parts.
In this task, the given piecewise function \(f(x)\) switches equations at the point \(x=2\). Such functions are practically useful, but they require careful analysis for continuity and differentiability, especially at the transition points:

• Identify each piece and determine the domain segment it covers.
• Investigate behavior at the transition points through limits, ensuring matching values for continuity.
• Check the existence and equality of derivatives at transition points to confirm differentiability.

By analyzing each piece separately and ensuring smooth transitions at the join points, we ascertain the overall function's compliance with rules like the Mean Value Theorem. This exercise not only demonstrates how piecewise functions can model diverse mathematical behaviors but also underlines the importance of meticulous checking for applying calculus principles effectively.