SOLUTION 2: We are given the function $f(x) = x^2(x-1) = x^3-x^2$ and the interval $[0, 3]$. This function is continuous on the closed interval $[0, 3]$ since it is a polynomial. The derivative of $f$ is $$f'(x) = 3x^2-2x$$ We can now see that $f$ is differentiable on the open interval $(0, 3)$. The assumptions of the Mean Value Theorem have now been met. Let's apply the Mean Value Theorem and find all possible values of $c$ in the open interval $(0, 3)$. Then $$f'(c)= \displaystyle{ f(3)-f(0) \over 3-0 } \ \ \ \ \longrightarrow$$ $$3c^2-2c = \displaystyle{ ( (3)^3 - (3)^2) ) - ( (0)^3 - (0)^2 ) \over 3-0 } \ \ \ \ \longrightarrow$$ $$3c^2-2c = \displaystyle{ 18 - 0 \over 3 } \ \ \ \ \longrightarrow$$ $$3c^2-2c = 6 \ \ \ \ \longrightarrow$$ $$3c^2-2c-6 = 0 \ \ \ \ \longrightarrow$$ $$(3)c^2+(-2)c+(-6) = 0 \ \ \ \ \longrightarrow$$ (Now use the Quadratic Formula.) $$c = \displaystyle{ -(-2) \pm \sqrt{ (-2)^2 - 4(3)(-6) } \over 2(3) } \ \ \ \ \longrightarrow$$ $$c = \displaystyle{ 2 \pm \sqrt{ 76 \ } \over 6 } \ \ \ \ \longrightarrow$$ $$c = \displaystyle{ 2 \pm 2 \sqrt{ 19 \ } \over 6 } \ \ \ \ \longrightarrow$$ $$c = \displaystyle{ 1 \pm \sqrt{ 19 \ } \over 3 } \ \ \ \ \longrightarrow$$ $$c = \displaystyle{ 1 + \sqrt{ 19 \ } \over 3 } \approx 1.786 \ \ \ or \ \ \ \displaystyle{ 1 - \sqrt{ 19 \ } \over 3 } \approx -1.120 \ \ \ (-1.120 \ is \ NOT \ in \ the \ interval \ (0, 3). ) \ \ \ \ \longrightarrow$$ $$c = \displaystyle{ 1 + \sqrt{ 19 \ } \over 3 } \approx 1.786$$