Question
Question: If f(x) = x<sup>3</sup> + ax<sup>2</sup> + bx + c attains it's local minima at certain negative real...
If f(x) = x3 + ax2 + bx + c attains it's local minima at certain negative real number then
A
a2− 3b > 0, a < 0, b < 0
B
a2− 3b > 0, a < 0, b > 0
C
a2 – 3b > 0, a > 0, b < 0
D
a2 – 3b > 0, a > 0, b > 0
Answer
a2 – 3b > 0, a > 0, b > 0
Explanation
Solution
f '(x) = 3x2 + 2ax + b = 3 (x – x1) (x − x2) where x1 < x2. clearly f '(x) < 0 ∀ x ∈ (x1, x2) and f '(x) > 0 ∀ x ∈ (-∞, x1) ∪ (x2, ∞). Thus x = x1 is the point of local maxima and x = x2 is the point of local minima.
Thus bigger root of f '(x) = 0 must be negative. Hence
a2 − 3b > 0, a > 0, b > 0