What-is-the-cubic-polynomial-for-y-0-1-y-1-0-y-2-1-and-y-3-10-

Question Number 105735 by bobhans last updated on 31/Jul/20

W h a t i s t h e c u b i c p o l y n o m i a l f o r y (0) = 1;

y (1) = 0; y (2) = 1 a n d y (3) = 10

Commented by PRITHWISH SEN 2 last updated on 31/Jul/20

y (x) = (x - 1) ({ax}^{2} + bx - 1)

y (2) = 4 a + 2 b - 1 = 1 \Rightarrow 2 a + b = 1

y (3) = 2 (9 a + 3 b - 1) = 10 \Rightarrow 3 a + b = 2

a = 1 b = - 1

y (x) = (x - 1) (x^{2} - x - 1) = x^{3} - 2 x^{2} + 1

Answered by bemath last updated on 31/Jul/20

L e t y (x) = (x - 1) {(x - 1) + a x (x - 2)}

a p l l y i n g t h e g i v e n c o n d i t i o n

y (3) = 2 . {2 + 3 a (1)} = 10

2 + 3 a = 5 \Rightarrow a = 1

y (x) = (x - 1) {(x - 1) + x (x - 2)}

y (x) = (x - 1) {x^{2} - x - 1}

y (x) = x^{3} - 2 x^{2} + 1 . ★

v e r i f i c a t i o n

y (2) = 8 - 8 + 1 = 1

y (0) = 1

y (1) = 1 - 2 + 1 = 0

y (3) = 27 - 18 + 1 = 10

Answered by floor(10²Eta[1]) last updated on 31/Jul/20

y (0) = 1 \Rightarrow y (x) = a (x) (x) + 1

y (1) = a (1) + 1 = 0 \Rightarrow a (1) = - 1

\Rightarrow a (x) = b (x) (x - 1) - 1

\Rightarrow y (x) = [b (x) (x - 1) - 1] x + 1

y (2) = [b (2) - 1] . 2 + 1 = 1 \Rightarrow b (2) = 1

\Rightarrow b (x) = c (x) (x - 2) + 1

\Rightarrow y (x) = [[c (x) (x - 2) + 1] . (x - 1) - 1] x + 1

y (3) = [[c (3) + 1] . 2 - 1] . 3 + 1 = 10 \Rightarrow c (3) = 1

\Rightarrow c (x) = 1 (constant bc y (x) is cubic)

\Rightarrow y (x) = [(x - 1) (x - 1) - 1] . x + 1

y (x) = [x^{2} - 2 x] . x + 1

\Rightarrow y (x) = x^{3} - {2 x}^{2} + 1