Posted by Joey on Monday, November 5, 2012 at 7:11pm.
0 = f(1) = a+b+c+d
2 = f(-2) = -8a+4b-2c+d
0 = f'(1) = 3a+2b+c
0 = f'(-2) = 12a-4b+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!
- Differential Calc - Steve, Monday, November 5, 2012 at 9:39pm
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 = 12a-4b
6b = 9a
2b = 3a
take the first two equations, equate d:
a+b+c = -8a+4b-2c
2a+2b+2c = -16a+8b-4c
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(x-1)^2 (2x+7)
f(1) = 0 and f(-2) = 2
y' = 4/9 (x^2 + x - 2) = 4/9 (x+2)(x-1)
it's easy to see that f'(1) = f'(-2) = 0
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