Mean Value Theorem Solution 10

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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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