Question 2.10.3: An Illustration of the Mean Value Theorem Find a value of c ...

Notice that f is continuous on [0, 2] and differentiable on (0, 2). The Mean
Value Theorem then says that there is a number c in (0, 2) for which

To find this number c, we set

or 3 c^2-2 c-2=0.

From the quadratic formula, we get c=\frac{1 \pm \sqrt{7}}{3}. In this case, only one of these,

c=\frac{1+\sqrt{7}}{3},is in the interval (0, 2). In Figure 2.53, we show the graphs of y = f (x),

the secant line joining the endpoints of the portion of the curve on the interval [0, 2] and the tangent line at x=\frac{1+\sqrt{7}}{3}.