Differential Calc
posted by Joey .
If:
0 = f(1) = a+b+c+d
2 = f(2) = 8a+4b2c+d
0 = f'(1) = 3a+2b+c
0 = f'(2) = 12a4b+c
"since we now have four equations and four unknowns, we can determine unique solution. Solving for "a" and "b", a=, b="
Please show the steps, I am so confused and no one knows how to do it.
Thank you so much in advance!

apparently f(x) = ax^3 + bx^2 + cx + d
so f'(x) = 3ax^2 + 2bx + c
and you were given f(1), f(2) and f'(1), f'(2)
take the last two equations, equate c:
3a+2b = 12a4b
6b = 9a
2b = 3a
take the first two equations, equate d:
a+b+c = 8a+4b2c
so,
2a+2b+2c = 16a+8b4c
but 2b = 3a, so
2a+3a+c = 16a + 12a  4c
9a = 5c
start substituting those values back in, and you get
a,b,c,d = 4/27, 2/9, 8/9, 14/27
y = 1/27 (4x^3 + 6x^2  24x + 14)
= 2/27(x1)^2 (2x+7)
f(1) = 0 and f(2) = 2
y' = 4/9 (x^2 + x  2) = 4/9 (x+2)(x1)
it's easy to see that f'(1) = f'(2) = 0