Question 196447
x,,,,,,1,,,,,2,,,,3,,,,,,4,,,,,5,,,,,,6
.
f(x),,,,0,,,10,,,24...56,,,112,,,190,,,,,,,,base
.
d,,,,,,,,,,10,,,,14,,,32,,,56,,,78,,,,,,,,,,,,,1st degree
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d,,,,,,,,,,,,,,,,,4,,,,,18,,,24,,,22,,,,,,,,,,,,,,2nd  degree
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d,,,,,,,,,,,,,,,,,,,,,14,,,,6,,,,,-2,,,,,,,,,,,,,,,,,3rd  degree
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d,,,,,,,,,,,,,,,,,,,,,,,,,,-8,,,,-8,,,,,,,,,,,,,,,,,,,4th  degree
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This  defines  f(x)  =  ax^4  +bx^3 + cx^2 + dx +e
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(1)  0 = a(1)^4  +b(1)^3 +c(1)^2 +d(1)^1 +e
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(2) 10 = a(2)^4 +b(2)^3 +c(2)^2 +d(2)^1 +e
.
(3) 24 = a(3)^4  +b(3)^3 +c(3)^2 +d(3)^1 +e
.
(4) 56 = a(4)^4 +b(4)^3 +c(4)^2 +d(4)^1 +e
.
(5) 112  =a(5)^4 +b(5)^3  +c(5)^2 +d(5)^1 +e
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(6) 190 = a(6)^4 +b(6)^3 +c(6)^2 +d(6)^1 +e
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now  solving  these  simultaneously,  using successive  substitutions (and  mult),  
or  matrix  computation
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Results  in approx,,, f(x) = -.03 x^4 +1.49x^3 -7.32x^2 +21.93x-16.07
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checking  shows  good  agreement  at  low  end,   but  about  70 %  of  actual  at  high  end
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making  this  correction,,,f(7) = 311  approx
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Time  does  not  permit any  further  computation, but  perhaps  this  will  give  you  a  good  start