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SOLUTION 10: We are given the function f(x)=(8x)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=8x (polynomial) and y=x1/3 (well-known). The derivative of f is f(x)=(1/3)(8x)2/3(1)=13(8x)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(8c)2/3=(88)1/3(80)1/38     13(8c)2/3=(0)1/3(8)1/38     13(8c)2/3=028     13(8c)2/3=14     3(8c)2/3=4     (8c)2/3=43     ((8c)2)1/3=43     (8c)2=(43)3     (8c)2=6427     8c=±6427     8c=±83 3     c=883 33.381   or   c=8+83 312.619 (12.619 is NOT in the interval (0,8).)     c=883 3

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