Question 35062: My book doesn’t want to talk about it, they just give you the answer and that’s hopeless. Please what are the operations to do here I just can’t find it.
A polynomial f(t) of degree 3 satisfies f(−2) = 0, f(2) = 4, f'(−1) = 2, and f'(1) = −6.
(a) Write the general form of the polynomial and its derivative.
(b) Find four linear equations that must be satisfied by the unknown coefficients in the
polynomial.
(c) Write the augmented matrix for your system of equations in (b), row-reduce by hand
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! I have been on this for hours, if any one can please direct me.
A polynomial f(t) of degree 3 satisfies f(−2) = 0, f(2) = 4, f'(−1) = 2, and f'(1) = −6.
(a) Write the general form of the polynomial and its derivative.
(b) Find four linear equations that must be satisfied by the unknown coefficients in the polynomial.
(c) Write the augmented matrix for your system of equations in (b) and row-reduce.
THE GENERAL FORM OF POLYNOMIAL OF DEGREE 3 IS
F(T)= AT^3+BT^2+CT+D ....................EQN.A
DF/DT= 3AT^2+2BT+C...................EQN.B
WE ARE GIVEN F(-2)=0...SO SUBSTITUTING IN EQN.A
-8A+4B-2C+D=0.............................EQN.I
F(2)=4...SO SUBSTITUTING IN EQN.A
8A+4B+2C+D=4................................EQN.II
F'(-1)=2...SO SUBSTITUTING IN EQN.B
3A-2B+C=2......................................EQN.III
F'(1)=-6...SO SUBSTITUTING IN EQN.B
3A+2B+C=-6..................................EQN.IV
AUGMENTED MATRIX A IS
-8,4,-2,1,0
8,4,2,1,4
3,-2,1,0,2
3,2,1,0,-6
I HOPE YOU CAN REDUCE THIS NOW AS YOU WANTED DIRECTION ...IF IN DOUBT SEE FOLLOWING EXAMPLE TO KNOW THE METHOD.
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The problem:
2x-5y+ ___=-22
__+y+3z=10
x+___+8z=15
The underscores represents blanks or O
If this had to be an augmented matrix, how would I do that and get those variables?
1 solutions
Answer 20202 by venugopalramana(1462) About Me on 2006-04-16 22:01:33 (Show Source):
2x-5y+ ___=-22...........................I
__+y+3z=10.....................II
x+___+8z=15...........................III
2*EQN.III-EQN.I
2X+16Z-2X+5Y=30+22=52
5Y+16Z=52...........................IV
EQN.IV-5*EQN.II
5Y+16Z-5Y-15Z=52-50=2
Z=2....SUBSTITUTING IN EQN.III
X+8*2=15
X=15-16=-1
SUBSTITUTING FOR Z IN EQN.II
Y+3*2=10
Y=10-6=4
USING MATRIX METHOD...AUGMENTED MATRIX IS
2 -5 0 -22 1 0 0 ?
0 1 3 10 0 1 0 ?
1 0 8 15 0 0 1 ?
NR1=R1/2
1 -2.5 0 -11
0 1 3 10
1 0 8 15
NR3=R3-R1
1 -2.5 0 -11
0 1 3 10
0 2.5 8 26
NR3=R3-2.5*R2
1 -2.5 0 -11
0 1 3 10
0 0 0.5 1
NR3=R3/0.5
1 -2.5 0 -11
0 1 3 10
0 0 1 2
NR2=R2-3R3
1 -2.5 0 -11
0 1 0 4
0 0 1 2
NR1=R1+2.5R2
1 0 0 -1
0 1 0 4
0 0 1 2
HENCE
X=-1
Y=4
Z=2
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