We have the cubic polynomial function f(x) = ax3+bx2 + cx + d f(1) = −1 = a + b + c + d f(2) = 18 = 8a + 4b + 2c + d 19 = 7a + 3b + c (1) Therefore, f ′(x) = 3ax2 + 2bx + c f ′(-1) = 0 3a – 2b + c = 0 3a + c = 0 c = −3a (2) and f ′′(x) = 6ax + 2b f ′′(0) = 0 b = 0 (3) Substituting Eqs. (2) and (3) in Eq. (1), we get 19 = 7a – 3a ⇒ a =
19
4
Therefore, c = −
57
4
and d = −1 – a – b – c = - 1 -
19
4
- 0 -
57
4
= -
(4+19+57)
4
= -
80
4
Therefore, f (x) =
19
4
x3−
57x
4
−
80
4
=
1
4
(19x3−57x−80) and f' (x) = 0 ⇒
1
4
(57x2−57) = 0 (x2 – 1) = 0 ⇒ x = 1, −1 f' (x) =
57
4
(x - 1) (x + 1) Therefore, f" (x) = (
57
2
)2x ⇒ f ′′(1) > 0 The point of minima is x = 1. Therefore, f (1) =
1
4
(19 - 57 - 80) =
59
2
=
−59
2
Point (1,
−59
2
) distance from point (−1, 2) is √(1+1)2(
−59
2
−2)2 = √4+
(63)2
4
=
1
2
√632+42 Hence, f(x) is increasing, that is, x ≥ 1.