Question

suppose ( f ) is known to be differentiable and contains a local extremum at ( c ), then which of the following is true?

( \bigcirc f(c) ) does not exist
( \bigcirc f(c)<0 )
( \bigcirc f(c)=0 )

Explanation:

Step1: Recall Fermat's Theorem

Fermat's Theorem states that if a function \(y = f(x)\) is differentiable at a point \(x = c\) and \(f(x)\) has a local extremum at \(x = c\), then \(f^{\prime}(c)=0\).

Step2: Analyze the given options

• The first option \(f^{\prime}(c)\) does not exist: But the function is given to be differentiable at \(c\), so \(f^{\prime}(c)\) exists.
• The second option \(f^{\prime}(c)<0\): If \(f^{\prime}(c)<0\), the function is decreasing at \(x = c\), which does not imply a local extremum.
• The third option \(f^{\prime}(c) = 0\): By Fermat's Theorem, since \(f\) is differentiable at \(c\) and has a local extremum at \(c\), \(f^{\prime}(c)=0\).

Answer:

\(f^{\prime}(c)=0\)