Using Rolle’s Theorem

Find the \(x\)-intercepts of \(f(x) = x^{2} - 5x + 6\), and show that \(f^\prime (c) = 0\) for some number \(c\) belonging to the interval formed by the two \(x\)-intercepts. Find \(c\).

Solution At the \(x\)-intercepts, \(f(x) = 0\). \[f(x)=x^{2}-5x+6=({x-2})({x-3})=0\]

So, \(x=2\) and \(x=3\) are the \(x\)-intercepts of the graph of \(f\), and \(f(2) =\) \(f(3) =0\).

Since \(f\) is a polynomial, it is continuous on the closed interval \([2,3]\) formed by the \(x\)-intercepts and is differentiable on the open interval \((2,3)\). The three conditions of Rolle’s Theorem are satisfied, guaranteeing that there is a number \(c\) in the open interval \((2,3)\) for which \(f^\prime (c) = 0\). Since \(f^\prime (x) = 2x-5,\) the number \(c\) for which \(f^\prime (x) =0\) is \(c=\dfrac{5}{2}.\)

See Figure 23 for the graph of \(f\).