Problem 14 - University Calculus: Early Transcendentals

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. $$f(x)=\left\\{\begin{array}{ll}2 x-3, & 0 \leq x \leq 2 \\ 6 x-x^{2}-7, & 2 < x \leq 3\end{array}\right.$$

Step-by-Step Solution

Answer

The function satisfies the Mean Value Theorem on the interval \([0, 3]\).

Step 1: Understand the Mean Value Theorem

The Mean Value Theorem (MVT) states that if a function \( f(x) \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). We need to check these conditions for \( f(x) \) on the interval \([0, 3]\).

Step 2: Check Continuity on [0, 3]

Analyze the function piecewise. For \( 0 \leq x \leq 2 \), \( f(x) = 2x - 3 \) is a linear function, hence continuous. For \( 2 < x \leq 3 \), \( f(x) = 6x - x^2 - 7 \), a quadratic function, is also continuous. Check the value of \( f(x) \) at \( x = 2 \): the left limit is \( f(2) = 2(2) - 3 = 1 \) and the right limit is also \( f(2) \) from the quadratic side, which yields \( 6(2) - 2^2 - 7 = 1 \). Thus, \( f(x) \) is continuous on \([0, 3]\).

Step 3: Check Differentiability on (0, 3)

For \( 0 < x < 2 \), the derivative \( f'(x) = 2 \) exists because it's a linear function. For \( 2 < x < 3 \), the derivative is \( f'(x) = 6 - 2x \), a continuous function. Check differentiability exactly at \( x=2 \): the derivative from the left is \( 2 \) and from the right at a point infinitely close, the derivative would be \( 6 - 2(2) = 2 \). Therefore, the function is differentiable on \((0, 3)\).

Step 4: Conclusion of MVT Application

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

Key Concepts

Continuity

For any function to satisfy the Mean Value Theorem (MVT), it must be continuous on the closed interval \[a, b\]. To determine continuity, we need to ensure that there are no breaks, jumps, or holes in the graph of the function over that interval.
For the function given in the exercise, it is defined piecewise as:

• For \([0, 2]\), the function is \(f(x) = 2x - 3\), which is linear and therefore continuous.
• For \((2, 3]\), the function is \(f(x) = 6x - x^2 - 7\), which is quadratic and inherently continuous.

To determine continuity at the point where the pieces connect, \(x = 2\), we evaluate the limit from the left side:
\[ \lim_{{x \to 2^-}} f(x) = f(2) = 2(2) - 3 = 1 \]
And from the right side using the second function definition:
\[ \lim_{{x \to 2^+}} f(x) = 6(2) - 2^2 - 7 = 1 \]
As both limits evaluate to the same value, the function is continuous at \(x = 2\), ensuring that \(f(x)\) is continuous across the whole interval \[0, 3\].
This confirms that the hypothesis of continuity required by the MVT is satisfied.

Differentiability

Differentiability is another crucial requirement for a function to satisfy the Mean Value Theorem. A function is said to be differentiable on an open interval \(a, b\) if it has a derivative at every point within that interval. To check differentiability for piecewise functions, examine each piece separately and their junctions.
For \(0 < x < 2\):

• The function is \(f(x) = 2x - 3\). Its derivative, \(f'(x) = 2\), exists as it is a constant, confirming differentiability on the interval.

For \(2 < x < 3\):

• The derivative is calculated as \(f'(x) = 6 - 2x\), which exists for all \(x\) in this range as it is a linear expression.

Finally, at \(x = 2\), it is essential to compare the derivatives from each side:

• The left-hand derivative as \(x\) approaches \(2\) is \(f'(2^-) = 2\).
• The right-hand derivative as \(x\) approaches \(2\) is also \(f'(2^+) = 2\), calculated from \(f'(x) = 6 - 2x\).

Since the derivatives match, the function is differentiable at \(x = 2\) as well.
Therefore, \(f(x)\) satisfies the differentiability condition of the MVT on the interval \(0, 3\), as it is differentiable everywhere except possibly endpoints, which are not considered in this context.

Piecewise Functions

Piecewise functions are functions that have different expressions for different intervals of their domain. The given function in the exercise is composed of two different expressions: one linear and one quadratic, each applicable over specific parts of the domain. Understanding how these functions work is key when applying concepts like continuity and differentiability.
When dealing with piecewise functions:

• Consider each segment separately for analysis.
• Check if each piece is continuous and differentiable on its respective interval.
• Ensure that at their boundary points, such as \(x = 2\), each piece smoothly connects, without jumps or breaks both in value and in slope.

This function, while distinct in its segments, connects smoothly at \(x = 2\), with both function values and slopes matching. This seamless connection at interfaces is crucial for meeting the mean value theorem's criteria on continuity and differentiability. By clearly identifying these properties, it simplifies complex analysis and understanding of piecewise functions in mathematical contexts.