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find the midpoint of ab, where
a(7,6) and b(4, 2).

Suppose that f(0) = −3 and f '(x) ≤ 8 for all values of x. how large can f(4) possibly be? solution we are given that f is differentiable (and therefore continuous) everywhere. in particular, we can apply the mean value theorem on the interval [0, 4] . there exists a number c such that

Which of the following represents the zeros of f(x) = 6x3 − 31x2 + 4x + 5? −5, one third , one half 5, − one third , one half 5, one third , − one half 5, one third , one half

5(-9w) put expression in distributive property