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Chapter 3: Problem 53

Let \(f\) be continuous on \([a, b]\) and differentiable on \((a, b) .\) If there exists \(c\) in \((a, b)\) such that \(f^{\prime}(c)=0,\) does it follow that \(f(a)=f(b) ?\) Explain.

Short Answer

Expert verified
{\(f(a) = f(b)\)} only if there exists a \(c\) in \((a, b)\) such that \(f'(c) = 0\) due to Mean Value Theorem.

Step by step solution

01

Interpret the Problem

The problem states that there exists a \(c\) in \((a, b)\) such that the derivative of the function at \(c\), \(f'(c)\), is zero. We are asked to determine whether this implies \(f(a) = f(b)\)
02

Apply Mean Value Theorem (MVT)

Due to MVT, because \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\), there exists a \(c\) in \((a, b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). Here, we are given that \(f'(c)=0\). Therefore, it equates to \(0 = \frac{f(b) - f(a)}{b - a}\).
03

Extract the result from the equation

Solving the above equation, we achieve that \(f(b) - f(a) = 0\), implying \(f(b) = f(a)\). Hence, if there exists a point \(c\) in \((a, b)\) such that \(f'(c)=0\), then indeed it follows that \(f(a) = f(b)\).

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