Question

analyzing tables determining a local maximum and minimum analyze the table of values for the continuous function, f(x), to complete the statements. a local maximum occurs over the interval a local minimum occurs over the interval

Explanation:

Step1: Recall local - max/min definition

A local maximum of a function \(y = f(x)\) occurs at a point \(x = c\) if \(f(c)\geq f(x)\) for all \(x\) in some open interval containing \(c\), and a local minimum occurs at a point \(x = c\) if \(f(c)\leq f(x)\) for all \(x\) in some open interval containing \(c\).

Step2: Analyze the table values

Looking at the table of values for \(x=-3,-2,-1,0,1,2\) and corresponding \(f(x)= - 16,-1,2,-1,-4,-1\).
We see that \(f(-1) = 2\). For \(x=-2\), \(f(-2)=-1\) and for \(x = 0\), \(f(0)=-1\). Since \(f(-1)=2\) is greater than the values of \(f(x)\) in its immediate - neighborhood (\(x=-2\) and \(x = 0\)), a local maximum occurs at \(x=-1\).
We also see that \(f(-2)=-1\). For \(x=-3\), \(f(-3)=-16\) and for \(x=-1\), \(f(-1)=2\). Since \(f(-2)=-1\) is less than the value of \(f(x)\) at \(x=-1\) and greater than the value of \(f(x)\) at \(x=-3\), a local minimum occurs at \(x=-2\).

Answer:

A local minimum occurs over the interval \(x=-3\) to \(x=-1\) at \(x = - 2\). A local maximum occurs over the interval \(x=-2\) to \(x = 0\) at \(x=-1\).