Let (2 , ) be the largest interval in which the function f(t) = |t+1|t2, t < 0, is strictly decreasing. Then the local maximum value of the function g(x) = 2 e(x-2) + x2 + 4x - , x > 2, is
Answer
4
Solution
For -1 < t < 0: f(t) = t+1t2 f'(t) = -(t+2)t3 > 0 (increasing)
For t < -1: f(t) = -(t+1)t2 f'(t) = t+2t3
f'(t) < 0 when t + 2 > 0 (as t3 < 0), i.e. -2 < t < -1
f is strictly decreasing on (-2, -1) = (2 , ) = -1
g(x) = 2 e(x-2) - x2 + 4x + 1, x > 2
g'(x) = 2x-2 - 2x + 4 = 2x-2 - 2(x-2)
g'(x) = 0 (x-2)2 = 1 x = 3 (as x > 2)
g''(x) = - 2(x-2)2 - 2 < 0 local maximum at x = 3
g(3) = 2 e 1 - 9 + 12 + 1 = 4
Answer: 4