Problem 27 - Calculus Early Transcendentals

Question

(a) Use the Mean-Value Theorem to show that if \(f\) is differentiable on an interval, and if \(\left|f^{\prime}(x)\right| \leq M\) for all values of \(x\) in the interval, then $$ |f(x)-f(y)| \leq M|x-y| $$ for all values of \(x\) and \(y\) in the interval. (b) Use the result in part (a) to show that $$ |\sin x-\sin y| \leq|x-y| $$ for all real values of \(x\) and \(y .\)

Step-by-Step Solution

Answer

(a) The MVT implies \(|f(x)-f(y)| \leq M|x-y|\). (b) For \(f(x)=\sin x\), \(|\sin x-\sin y| \leq |x-y|\) holds.

Step 1: Understanding the Mean Value Theorem (MVT)

The Mean Value Theorem states: If a function \( f \) is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \( c \) in \( (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). This theorem provides a way to relate the difference in the values of \( f \) at two points to its derivative.

Step 2: Applying MVT to \(f(x)\) and \(f(y)\)

According to MVT, if \(f\) is differentiable on an interval, for any two points \(x\) and \(y\) in this interval, there exists some point \(c\) between \(x\) and \(y\) such that: \[ f'(c) = \frac{f(x) - f(y)}{x - y} \] Multiplying both sides by \(|x-y|\) gives: \[ |f(x) - f(y)| = |f'(c)| \cdot |x-y| \] Since \(|f'(x)| \leq M\), this implies \(|f'(c)| \leq M\). Thus, we have: \[ |f(x) - f(y)| \leq M|x-y| \]

Step 3: Verifying the Result for \(f(x) = \sin x\)

For \( f(x) = \sin x \), the derivative is \( f'(x) = \cos x \). Since \( |\cos x| \leq 1 \) for all real numbers \( x \), we can take \( M = 1 \). Applying the result from part (a), we have for any real \( x \) and \( y \):\[ |\sin x - \sin y| \leq 1 \cdot |x-y| \]Thus, \( |\sin x - \sin y| \leq |x-y| \) for all real \( x \) and \( y \).

Key Concepts

Differentiable functions

Differentiable functions are at the heart of calculus, playing a crucial role in understanding real-world problems. A function is differentiable over an interval if it has a derivative at every point within that interval. In simpler terms, it means you can find the slope of the tangent to the function’s graph at any point. This concept is essential for applying the Mean Value Theorem (MVT), which relies on differentiability to establish a relationship between function values and their derivatives.

When a function is differentiable, it is also continuous. However, the converse isn't necessarily true; a continuous function might not be differentiable. Differentiability implies that the function's graph is smooth, without any sharp "jumps" or "corners." This smoothness is crucial for calculating changes and understanding rates, making differentiable functions vital in fields like physics and engineering. Always remember: if a function is differentiable over an interval, it must also be continuous over that interval, serving as a smooth path to explore various function behaviors.

Derivative bounds

Derivative bounds give us constraints on how much a function can change. These bounds are based on the derivative's values across an interval. The concept is critical when applying the MVT, as it allows us to predict the possible maximum increase or decrease of a function over an interval.

A derivative bound, often denoted by the letter \( M \), means that for a function \( f \), the derivative at any point \( x \) in the interval satisfies \( |f'(x)| \leq M \). This bound can help in proving statements about the function's values. For example, in our original exercise, this concept allows us to conclude that \(|f(x) - f(y)| \leq M|x-y|\).

• This inequality shows how the function's value does not change wildly, as it's confined by the derivative bound \(M\).
• Thus, when a derivative is bounded, analyzing functions becomes more manageable because we can predict their behavior within the given limits.

Trigonometric functions

Trigonometric functions, like sine and cosine, are periodic functions extensively used in mathematics to model wave-like phenomena. Among these, \( \sin x \) and \( \cos x \) are particularly famous for their boundaries, which are straightforward and handy in various calculations.

When dealing with the exercise, we exploit the fact that the derivative of \( \sin x \) is \( \cos x \), and \( |\cos x| \leq 1 \) for all real numbers \( x \). This derivative limit ensures that the sine function alternates between -1 and 1, providing a helpful bound for proving inequalities involving \( \sin x \).

• The periodic nature of trigonometric functions means they repeat after certain intervals, allowing prediction of their values.
• Understanding these functions and their behaviors, like how they relate to each other through derivatives, is fundamental for accurately solving problems in oscillation and wave theory.

Continuous functions

Continuous functions appear everywhere in calculus, providing a smooth and unbroken graph over intervals. A function is continuous if, intuitively, you can sketch its graph without lifting your pen from the paper. For continuous functions, no sudden jumps or breaks occur, assuring a smooth connection between points.

The Mean Value Theorem requires functions to be continuous on closed intervals and differentiable on open intervals. This requirement ensures that there is a predictable behavior between any two points, a concept vital in mathematical modeling and real-world applications.

• Being continuous implies stability, meaning small changes in input cause only small changes in output.
• Continuity forms the foundation for analyzing limits and solving equations in calculus, as you can always expect a continuous path of data.