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
$$ f'(c) = \displaystyle{ f(8)-f(0) \over 8 - (0) } \ \ \ \ \longrightarrow $$
$$ \displaystyle{ -1 \over 3(8-c)^{2/3} } = \displaystyle{ (8-8)^{1/3} - (8-0)^{1/3} \over 8 } \ \ \ \ \longrightarrow $$
$$ \displaystyle{ -1 \over 3(8-c)^{2/3} } = \displaystyle{ (0)^{1/3} - (8)^{1/3} \over 8 } \ \ \ \ \longrightarrow $$
$$ \displaystyle{ -1 \over 3(8-c)^{2/3} } = \displaystyle{ 0 - 2 \over 8 } \ \ \ \ \longrightarrow $$
$$ \displaystyle{ -1 \over 3(8-c)^{2/3} } = \displaystyle{ -1 \over 4 } \ \ \ \ \longrightarrow $$
$$ 3(8-c)^{2/3} = 4 \ \ \ \ \longrightarrow $$
$$ (8-c)^{2/3} = \displaystyle{ 4 \over 3 } \ \ \ \ \longrightarrow $$
$$ ((8-c)^2)^{1/3} = \displaystyle{ 4 \over 3 } \ \ \ \ \longrightarrow $$
$$ (8-c)^2 = \Big(\displaystyle{ 4 \over 3 }\Big)^3 \ \ \ \ \longrightarrow $$
$$ (8-c)^2 = \displaystyle{ 64 \over 27 } \ \ \ \ \longrightarrow $$
$$ 8-c = \pm \sqrt{\displaystyle{ 64 \over 27 } } \ \ \ \ \longrightarrow $$
$$ 8-c = \pm \displaystyle{ 8 \over 3 \ \sqrt{3} } \ \ \ \ \longrightarrow $$
$$ c = 8 - \displaystyle{ 8 \over 3 \ \sqrt{3} } \approx 3.381 \ \ \ or \ \ \ c = 8 + \displaystyle{ 8 \over 3 \ \sqrt{3} } \approx 12.619 \ (12.619 \ is \ NOT \ in \ the \ interval \ (0, 8).) \ \ \ \ \longrightarrow $$
$$ c = 8 - \displaystyle{ 8 \over 3 \ \sqrt{3} } $$
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