# calculus 1

find the cubic function,f(x)=ax^3+bx^2+cx^d , has local maximum value of 3, at x=-2 and local minimum value of 0, at x=1.

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1. f' = 3ax^2 + 2bx + c
f'=0 at x = -2
f'=0 at x = 1
sum of roots = -1
product of roots = -2
so, -2b/3a = 1, so 2b = -3a
c/3a = -2, so c = -6a

f(x) = ax^2 - 3a/2 x - 6ax + d

f(-2) = 3 and f(1) = 0, so

4a + 3a + 12a + d = 3
a - 3a/2 - 6a + d = 0

a=2/9 d=7/9, so

f(x) = 2/9 x^3 + 1/3 x^2 - 4/3 x + 7/9

= 2/9 (x^2 + 3x^2 - 12x + 7)

====================================
or, since we know we have a double root at x=1,

f(x) = a(x-k)(x-1)^2
f'(x) = a(x-1)^2 + 2a(x-k)(x-1)
= a(x-1)(x-1 + 2x-2k)
= a(x-1)(3x-2k-1)
we know there's a max at x = -2, so k-7/2, and

f(x) = a(x+7/2)(x-1)^2
Since f(-2) = 3,
a(3/2)(9) = 3
a = 2/9
f(x) = 2/9 (a+7/2)(x-1)^2
= 2/9 x^2 + 1/3 x^2 - 4/3 x + 7/9

see the graph and analysis at

http://www.wolframalpha.com/input/?i=2%2F9+%28x%2B7%2F2%29%28x-1%29^2

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posted by Steve

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