Questions: If f is continuous on [a, b], which of the following must be true? There is a c in [a, b] with I f(c)=0 II f'(c)=(f(b)-f(a))/(b-a) III f(c)=(∫ from a to b f(x) dx)/(b-a) (A) II only (B) III only (C) I and II only (D) II and III only (E) I,II, and III

Solution Steps

To determine which of the given statements must be true for a continuous function \( f \) on the interval \([a, b]\), we can use the following theorems from calculus:

1. Intermediate Value Theorem (IVT): If \( f \) is continuous on \([a, b]\) and \( f(a) \) and \( f(b) \) have opposite signs, then there exists a \( c \in [a, b] \) such that \( f(c) = 0 \).
2. Mean Value Theorem (MVT): If \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists a \( c \in (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
3. Average Value of a Function: If \( f \) is continuous on \([a, b]\), then there exists a \( c \in [a, b] \) such that \( f(c) = \frac{1}{b - a} \int_a^b f(x) \, dx \).

Given these theorems, we can analyze the statements:

• Statement I: This is not necessarily true unless \( f(a) \) and \( f(b) \) have opposite signs.
• Statement II: This is guaranteed by the Mean Value Theorem.
• Statement III: This is guaranteed by the Average Value of a Function.

Thus, the correct answer is (D) II and III only.

To determine if \( f(c) = 0 \) for some \( c \in [a, b] \), we use the Intermediate Value Theorem (IVT). The IVT states that if \( f \) is continuous on \([a, b]\) and \( f(a) \) and \( f(b) \) have opposite signs, then there exists a \( c \in [a, b] \) such that \( f(c) = 0 \). However, without additional information about the signs of \( f(a) \) and \( f(b) \), we cannot guarantee that \( f(c) = 0 \). Therefore, Statement I is not necessarily true.

To determine if \( f'(c) = \frac{f(b) - f(a)}{b - a} \) for some \( c \in (a, b) \), we use the Mean Value Theorem (MVT). The MVT states that if \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists a \( c \in (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). Since \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), Statement II is guaranteed to be true.

To determine if \( f(c) = \frac{1}{b - a} \int_a^b f(x) \, dx \) for some \( c \in [a, b] \), we use the concept of the average value of a function. The average value of a continuous function \( f \) on \([a, b]\) is given by \( \frac{1}{b - a} \int_a^b f(x) \, dx \). The Mean Value Theorem for integrals states that there exists a \( c \in [a, b] \) such that \( f(c) = \frac{1}{b - a} \int_a^b f(x) \, dx \). Therefore, Statement III is guaranteed to be true.

Final Answer