Use the Mean Value Theorem from §4.2 to prove that | sin(b) – sin(a)| < 16 – a| for any interval [a, b].

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Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Applying the Mean Value Theorem to Prove Inequality**

To prove the inequality:

\[ | \sin(b) - \sin(a) | \leq |b - a| \]

for any interval \([a, b]\), we will utilize the Mean Value Theorem (MVT) as discussed in Section 4.2.

### Mean Value Theorem Statement
The MVT states that if a function \(f\) is continuous on the 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:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

### Proof Applying the MVT
Consider the function \(f(x) = \sin(x)\), which is continuous and differentiable everywhere. By the MVT, there exists a point \(c \in (a, b)\) such that:

\[ \sin'(c) = \frac{\sin(b) - \sin(a)}{b - a} \]

We know that \(\sin'(x) = \cos(x)\). Therefore:

\[ \cos(c) = \frac{\sin(b) - \sin(a)}{b - a} \]

Taking the absolute value of both sides:

\[ |\cos(c)| = \left| \frac{\sin(b) - \sin(a)}{b - a} \right| \]

We also know that the maximum value of \(|\cos(x)|\) is 1 for all \(x\). Thus,

\[ |\cos(c)| \leq 1 \]

This implies:

\[ \left| \frac{\sin(b) - \sin(a)}{b - a} \right| \leq 1 \]

Multiplying through by \(|b - a|\), we get:

\[ |\sin(b) - \sin(a)| \leq |b - a| \]

which completes the proof.
Transcribed Image Text:**Applying the Mean Value Theorem to Prove Inequality** To prove the inequality: \[ | \sin(b) - \sin(a) | \leq |b - a| \] for any interval \([a, b]\), we will utilize the Mean Value Theorem (MVT) as discussed in Section 4.2. ### Mean Value Theorem Statement The MVT states that if a function \(f\) is continuous on the 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: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] ### Proof Applying the MVT Consider the function \(f(x) = \sin(x)\), which is continuous and differentiable everywhere. By the MVT, there exists a point \(c \in (a, b)\) such that: \[ \sin'(c) = \frac{\sin(b) - \sin(a)}{b - a} \] We know that \(\sin'(x) = \cos(x)\). Therefore: \[ \cos(c) = \frac{\sin(b) - \sin(a)}{b - a} \] Taking the absolute value of both sides: \[ |\cos(c)| = \left| \frac{\sin(b) - \sin(a)}{b - a} \right| \] We also know that the maximum value of \(|\cos(x)|\) is 1 for all \(x\). Thus, \[ |\cos(c)| \leq 1 \] This implies: \[ \left| \frac{\sin(b) - \sin(a)}{b - a} \right| \leq 1 \] Multiplying through by \(|b - a|\), we get: \[ |\sin(b) - \sin(a)| \leq |b - a| \] which completes the proof.
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