SOLUTION 10: We are given the function $f(x) = (8-x)^{1/3}$ and the interval $[0, 8]$. This function is continuous on the closed interval $[0, 8]$ since it is the functional composition of continuous functions $y=8-x$ (polynomial) and $y=x^{1/3}$ (well-known). The derivative of $f$ is $$f'(x) = (1/3)(8-x)^{-2/3} \cdot(-1) = \displaystyle{ -1 \over 3(8-x)^{2/3} }$$ We can now see that $f$ is differentiable on the open interval $(0, 8)$. 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 $(0, 8)$. Then $$f'(c) = \displaystyle{ f(8)-f(0) \over 8 - (0) } \ \ \ \ \longrightarrow$$ $$\displaystyle{ -1 \over 3(8-c)^{2/3} } = \displaystyle{ (8-8)^{1/3} - (8-0)^{1/3} \over 8 } \ \ \ \ \longrightarrow$$ $$\displaystyle{ -1 \over 3(8-c)^{2/3} } = \displaystyle{ (0)^{1/3} - (8)^{1/3} \over 8 } \ \ \ \ \longrightarrow$$ $$\displaystyle{ -1 \over 3(8-c)^{2/3} } = \displaystyle{ 0 - 2 \over 8 } \ \ \ \ \longrightarrow$$ $$\displaystyle{ -1 \over 3(8-c)^{2/3} } = \displaystyle{ -1 \over 4 } \ \ \ \ \longrightarrow$$ $$3(8-c)^{2/3} = 4 \ \ \ \ \longrightarrow$$ $$(8-c)^{2/3} = \displaystyle{ 4 \over 3 } \ \ \ \ \longrightarrow$$ $$((8-c)^2)^{1/3} = \displaystyle{ 4 \over 3 } \ \ \ \ \longrightarrow$$ $$(8-c)^2 = \Big(\displaystyle{ 4 \over 3 }\Big)^3 \ \ \ \ \longrightarrow$$ $$(8-c)^2 = \displaystyle{ 64 \over 27 } \ \ \ \ \longrightarrow$$ $$8-c = \pm \sqrt{\displaystyle{ 64 \over 27 } } \ \ \ \ \longrightarrow$$ $$8-c = \pm \displaystyle{ 8 \over 3 \ \sqrt{3} } \ \ \ \ \longrightarrow$$ $$c = 8 - \displaystyle{ 8 \over 3 \ \sqrt{3} } \approx 3.381 \ \ \ or \ \ \ c = 8 + \displaystyle{ 8 \over 3 \ \sqrt{3} } \approx 12.619 \ (12.619 \ is \ NOT \ in \ the \ interval \ (0, 8).) \ \ \ \ \longrightarrow$$ $$c = 8 - \displaystyle{ 8 \over 3 \ \sqrt{3} }$$