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 =
Therefore, c =
− and d = −1 – a – b – c
= - 1 -
- 0 -
= -
= -
Therefore, f (x) =
x3−− =
(19x3−57x−80) and f' (x) = 0 ⇒
(57x2−57) = 0
(
x2 – 1) = 0 ⇒ x = 1, −1
f' (x) =
(x - 1) (x + 1)
Therefore, f" (x) =
()2x ⇒ f ′′(1) > 0
The point of minima is x = 1. Therefore,
f (1) =
(19 - 57 - 80) =
=
Point
(1,) distance from point (−1, 2) is
√(1+1)2(−2)2 =
√4+ =
√632+42 Hence, f(x) is increasing, that is, x ≥ 1.