What is the cubic polynomial for y(0)=1; y(1)=0 ; y(2)=1 and y(3)=10


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
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
	$Let y(x)=(x−1){(x−1)+ax(x−2)} apllying the given condition y(3)=2.{2+3a(1)}=10 2+3a = 5 ⇒ a = 1 y(x)= (x−1){(x−1)+x(x−2)} y(x)=(x−1){x^2 −x−1} y(x)=x^3 −2x^2 +1 .★ verification y(2)=8−8+1=1 y(0) = 1 y(1)=1−2+1=0 y(3)=27−18+1=10$
	$L e t y (x) = (x - 1) {(x - 1) + a x (x - 2)}$ $a p l l y i n g t h e g i v e n c o n d i t i o n$ $y (3) = 2 . {2 + 3 a (1)} = 10$ $2 + 3 a = 5 \Rightarrow a = 1$ $y (x) = (x - 1) {(x - 1) + x (x - 2)}$ $y (x) = (x - 1) {x^{2} - x - 1}$ $y (x) = x^{3} - 2 x^{2} + 1 . ★$ $v e r i f i c a t i o n$ $y (2) = 8 - 8 + 1 = 1$ $y (0) = 1$ $y (1) = 1 - 2 + 1 = 0$ $y (3) = 27 - 18 + 1 = 10$
	Answered by floor(10²Eta[1]) last updated on 31/Jul/20

	$y (0) = 1 \Rightarrow y (x) = a (x) (x) + 1$ $y (1) = a (1) + 1 = 0 \Rightarrow a (1) = - 1$ $\Rightarrow a (x) = b (x) (x - 1) - 1$ $\Rightarrow y (x) = [b (x) (x - 1) - 1] x + 1$ $y (2) = [b (2) - 1] . 2 + 1 = 1 \Rightarrow b (2) = 1$ $\Rightarrow b (x) = c (x) (x - 2) + 1$ $\Rightarrow y (x) = [[c (x) (x - 2) + 1] . (x - 1) - 1] x + 1$ $y (3) = [[c (3) + 1] . 2 - 1] . 3 + 1 = 10 \Rightarrow c (3) = 1$ $\Rightarrow c (x) = 1 (constant bc y (x) is cubic)$ $\Rightarrow y (x) = [(x - 1) (x - 1) - 1] . x + 1$ $y (x) = [x^{2} - 2 x] . x + 1$ $\Rightarrow y (x) = x^{3} - {2 x}^{2} + 1$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum