SOLUTION: 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

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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
:
(1) -27a +9b -3c +d = -112
:
(2) -a +b -c +d = -2
:
(3) a +b +c +d = 4
:
(4) 8a +4b +2c +d = 13
:
Any of the following methods can be used, elimination method, substitution method, Gauss-Seidel method, Cramer's Rule
:
Note if you add equations 2 and 3 you get b +d = 1 and this can be used to eliminate a variable
:
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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