The average value of $f$ is $\frac{1}{3}{\int }_0^3\left(x^3-x\right) \mathit{\ⅆ}x$ = $\frac{21}{4}$.

The Mean Value theorem states that $f$ will attain its average value at least once on the interval $\left[0\,3\right]$. This occurs at the solution of the equation $f\left(x\right)\=21\/4$, which is

 $c\=\frac{1}{6}\frac{{\left(567\+3\sqrt{35529}\right)}^{2\/3}\+12}{{\left(567\+3\sqrt{35529}\right)}^{1\/3}}\doteq 1.929109400$ 

Since $\left(0\,f\left(0\right)\right)\=\left(0\,0\right)$ and $\left(3\,f\left(3\right)\right)\=\left(3\,24\right)$, the slope of the "secant" line is 8. The solution of the equation $f\prime \left(x\right)\=3 x^2-1\=8$ is $x\=\sqrt{3}$.

The point at which the function attains its average value is not the same point at which the tangent is parallel to the secant!

Figure 4.6.4(a) provides a screenshot of the  tutor applied to $f\left(x\right)$. Figure 4.6.4(b) provides a screen-shot of the  tutor applied to $f\left(x\right)$.

Table 4.6.4(a) details the calculations of the points where $f\left(x\right)\=f_{\mathrm{avg}}$ and where the tangent line is parallel to the secant line.

The most direct way to obtain the exact solution of the equation $f\left(x\right)\=f_{\mathrm{avg}}$ is shown below.

The default output for the MeanValueTheorem command is the graph shown in Figure 4.6.4(b). Alternate usages are shown in Table 4.6.4(b).