If f(x) is differentiable for the closed interval [-1, 4] such that f(-1) = -3 and f(4) = 12, then there exists a value c, -1 < c < 4 such that: A) f'(c) = 3 B) f'(c) = 0 C) f(c) = -15 D) f(c) = 3

  A)  f'(c) = 3

Step-by-step explanation:

The mean value theorem says that if f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists some c such that ...

  a < c < b

  f'(c) = (f(b) -f(a))/(b -a)

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We are told that f(x) is differentiable on the closed interval [-1, 4], so we know it meets the requirements of the mean value theorem. Then we can conclude that there is some c such that ...

  f'(c) = (12 -(-3))/(4 -(-1)) = 15/5

  f'(c) = 3 . . . . for some c in the interval -1 < c < 4

answer: just subtact the units and then add the non negative : happy to

step-by-step explanation: