Let the set of all positive values of , for which the point of local minimum of the function (1 + x ( 2 - x2 ) ) satisfies x2 + x + 2x2 + 5x + 6 < 0, be ( , ). Then 2 + 2 is equal to .
Answer
39
Solution
f(x) = 1 + 2 x - x3 f'(x) = 2 - 3x2 = 0 x = 3
f''(x) = -6x > 0 at x = - 3, so the local minimum is at x = - 3.
Since x2 + x + 2 > 0 always ( D = 1 - 8 < 0), we need (x + 2)(x + 3) < 0, i.e. -3 < x < -2:
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-3 < - 3 < -2 2 3 < < 3 3
So = 2 3, = 3 3 and 2 + 2 = 12 + 27 = 39.
Answer: 39