If f(x) is continuous and differentiable over [−2,5] and −4≤f′(x)≤3 for all x in (−2,5), then the greatest possible value of f(5)−f(−2) is
A
7
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B
9
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C
15
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D
21
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Solution
The correct option is D21 Given that −4≤f′(x)≤3 Using LMVT on f(x) in [−2,5], we get −4≤f(5)−f(−2)5−(−2)≤3 ⇒−28≤f(5)−f(−2)≤21 So, the greatest possible value of f(5)−f(−2) is 21.