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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=x1/3 (well-known). The derivative of f is
f′(x)=(1/3)(8−x)−2/3⋅(−1)=−13(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)=f(8)−f(0)8−(0) ⟶
−13(8−c)2/3=(8−8)1/3−(8−0)1/38 ⟶
−13(8−c)2/3=(0)1/3−(8)1/38 ⟶
−13(8−c)2/3=0−28 ⟶
−13(8−c)2/3=−14 ⟶
3(8−c)2/3=4 ⟶
(8−c)2/3=43 ⟶
((8−c)2)1/3=43 ⟶
(8−c)2=(43)3 ⟶
(8−c)2=6427 ⟶
8−c=±√6427 ⟶
8−c=±83 √3 ⟶
c=8−83 √3≈3.381 or c=8+83 √3≈12.619 (12.619 is NOT in the interval (0,8).) ⟶
c=8−83 √3
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