(1 point) Let pp be the quartic (degree 4) polynomial that satisfies p(i)=2i,i=0,1,2,3,4. p(i)=2i,i=0,1,2,3,4. Then p(x)=p(x)= .
Hint: You may have a better idea, but a brute force approach is to write p(x)=ax4+bx3+cx2+dx+e p(x)=ax4+bx3+cx2+dx+e where aa, bb, cc, dd, and ee, are the unknown coefficients, and then solve the linear system p(0)=1p(0)=1, p(1)=2p(1)=2, p(2)=4p(2)=4, p(3)=8p(3)=8, and p(4)=16p(4)=16 for aa, bb, cc, dd, and ee. Preview My AnswersSubmit Answers
1 answer:
Answer:
a = 1/3
b = -3
c = 26/3
d = -6
e = 0
Step-by-step explanation:
Given the quartic polynomial
p(x)=ax⁴+bx³+cx²+dx+e and
p(i) =2i when i=0,1,2,3,4
If i = 0:
p(0) = 2(0)
p(0) = 0
0 = 0+0+0+0+0++e
e = 0
When i = 1
p(1) = 2(1) = 2
2 = a(1)⁴+b(1)³+c(1)²+d(1)+e
2 = a+b+c+d+0
a+b+c+d = 0... (1)
When i = 2, p(2) = 2(2)
p(2) = 4
4 = a(2)⁴+b(2)³+c(2)²+d(2)+e
4 = 16a+8b+4c+2d+0
16a+8b+4c+2d = 4
8a+4b+2c+d = 2... (2)
When i = 3
p(3) = 8
8 = a(3)⁴+b(3)³+c(3)²+d(3)+0
8 = 81a+27b+9c+3d..(3)
When i = 4
p(4) =16
16 = a(4)⁴+b(4)³+c(4)²+d(4)+0
16 = 256a+64b+16c+4d
64a+16b+4c+d = 4...(4)
Solving equation 1 to 4 simultaneously.
Check the attachment for solution.
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