Answer:
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
