Apply Int. Value Theorem: f(x)= 2x+2/3x+1 on interval [0,1] & explain why number c exists such that f(c) = 5/3

f(x) is continuous on [0,1] since its only discontinuity is at x = -1/3





f(0) = 2


f(1) = 4/4 = 1





thus, there must exist some number c in [0,1] such that f(c) is between 2 and 1. Since 5/3 is between 2 and 1, there is a c sucht that f(c ) = 5/3

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