Question 1155338: Find the function f(x)=ax^3+bx^2+cx+d for which, f(-3)=-112, f(-1)=-2, f(1)=4, f(2)=13.
I know the answer is f(x) 3x^3-4x^2+5
But, i would like to know how to solve this
Found 2 solutions by rothauserc, MathTherapy: Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! There are 4 linear equations in 4 unknowns
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(1) -27a +9b -3c +d = -112
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(2) -a +b -c +d = -2
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(3) a +b +c +d = 4
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(4) 8a +4b +2c +d = 13
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Any of the following methods can be used, elimination method, substitution method, Gauss-Seidel method, Cramer's Rule
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Note if you add equations 2 and 3 you get b +d = 1 and this can be used to eliminate a variable
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Answer by MathTherapy(10552) (Show Source):
You can put this solution on YOUR website! Find the function f(x)=ax^3+bx^2+cx+d for which, f(-3)=-112, f(-1)=-2, f(1)=4, f(2)=13.
I know the answer is f(x) 3x^3-4x^2+5
But, i would like to know how to solve this
Given: f(- 3) = - 112, or (- 3, - 112)
Substituting (- 3, - 112) in , we get: ---- eq (i)
Given: f(- 1) = - 2, or (- 1, - 2)
Substituting (- 1, - 2) in , we get: ------ eq (ii)
Given: f(1) = 4, or (1, 4)
Substituting (1, 4) in , we get: ----------------- eq (iii)
Given: f(2) = 13, or (2, 13)
Substituting (2, 13) in , we get: ---------------- eq (iv)
We now have:
35A - 5B + 5C = 125____7A - B + C = 25 ---- Subtracting eq (i) from eq (iv) -------- eq (v)
7A + 3B + C = 9 ----- Subtracting eq (iii) from eq (iv) ------ eq (vi)
4B = - 16 ------ Subtracting eq (v) from eq (vi)
2B + 2D = 2____2(B + D) = 2(1) -- Adding eq (ii) & (iii)
B + D = 1 ------ eq (vii)
- 4 + D = 1 ------ Substituting - 4 for B in eq (vii)
A + B + C + D = 4
A - 4 + C + 5 = 4 -------- Substituting - 4 for B, and 5 for D in eq (iii)
A + C = 3 ------ eq (viii)
8A + 4B + 2C + D = 13 ------- eq (iv)
8A + 4(- 4) + 2C + 5 = 13 --- Substituting - 4 for B, and 5 for D in eq (iv)
8A - 16 + 2C + 5 = 13
8A + 2C = 24____2(4A + C) = 2(12)____4A + C = 12 ----- eq (ix)
3A = 9 ------ Subtracting eq (viii) from eq (ix)
A + C = 3 ----- eq (viii)
3 + C = 3 ----- Substituting 3 for A in eq (viii)
With , 
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