Question

True or False If \(m\) is a local minimum and \(M\) is a local maximum of a continuous function \(f\) on \((a, b),\) then \(m

Short answer

The statement is false. It is possible for \(m\) to be equal to \(M\), which would result in \(m<M\) being false.

Step-by-step solution

Understanding local minimum and maximum

A local minimum of a function is a point where the function attains a value which is less than or equal to the values of the function in a small interval around it. On the other hand, a local maximum of a function is a point where the function attains a value which is greater than or equal to the values of the function in a small interval around it.

Comparing local minimum and maximum

Recall that \((a, b)\) is an open interval, meaning \(a\) and \(b\) are not included in the interval. Since \(m\) is a local minimum, there are values of the function nearby that are greater or equal to \(m\). Similarly, since \(M\) is a local maximum, there are values of the function near \(M\) that are less or equal to \(M\).

Confronting the comparison with the question

The function is continuous and the given condition is that \(m\) is a local minimum and \(M\) is a local maximum. Since \(M\) can't be less than \(m\), because \(M\) is a local maximum and \(m\) is a local minimum, the statement \(m<M\) can be false only if \(m=M\). However, this doesn't violate the definition, as the definition of local extremum allows for the possibility that the extreme value is attained at several different points in its domain. Therefore, the statement isn't universally true.