Calculus AB
posted by Savannah .
Find a, b, c, and d such that the cubic function ax^3 + bx^2 + cx + d satisfies the given conditions
Relative maximum: (2,4)
Relative minimum: (4,2)
Inflection point: (3,3)
So this is what I have so far:
f'(x) = 3a^2 + 2bx + c
f''(x) = 6ax + 2b
f(2) = a(8) + b(4) + c(2) + d= 4
f(4) = a(64) + b(16) + c(4) + d = 2
> 30a + 6b + c = 1
Am I going in the right direction? If I am, what do I do next? Thanks.

You have a good start, but there are a few corrections to make:
f'(x) = 3a(x)^2 + 2bx + c
f''(x) = 6ax + 2b
f(2) = a(8) + b(4) + c(2) + d= 4
f(4) = a(64) + b(16) + c(4) + d = 2
> 30a + 6b + c = 1
You have three conditions for max. at (2,4):
f(2)=4
f'(2)=0
f"(2)<0
relative min. at f(4)=2 =>
f(4)=2
f'(4)=0
f"(4)>0
Inflexion point at (3,3) means:
f(3)=3
f"(3)=0
So you'd have six equations for four unknowns a,b, c and d. Notice that if the system is consistent (which is the case here), you only need 4 of the conditions to find the solutions, but you will have to check that the remaining conditions are satisfied.
Can you take it from here? 
your thinking is ok so far
also remember that (3,3) also lies on the original function, so
27a + 9b + 3c + d = 3 will be another equation.
How did you get 30a in 30a + 6b + c = 1
If you subtracted your two equations and then divided by 2 should it not have been
28a + 6b + c = 1 ? > #4
by subtracting my new equation from 64a +16b+4c+d = 2 , I get a new equation
37a + 7b+c=1 > # 5
#5  #4 :
9a + b = 0
b = 9a
also we know that f'(2) = 0 and f'(4) = 0
f'(2) = 12a + 4b + c = 0
f'(4) = 48a + 8b + c = 0
subtract:
36a + 4b = 0
b = 9a , nothing new here , confirms above.
but we also know f'(2) = f''(3) = 0
12a + 4b + c = 18a + 2b
c = 6a  2b
c = 6a  2(9a) = 24a
in #4:
28a + 6(9a) + 24a = 1
2a = 1
a = 1/2
b= 9/2
c = 12
so now back into : 8a + 4b + 2c + d = 4
4 18 +24 + d = 4
d = 6
so f(x) = (1/2)x^3 (9/2)x^2 + 12x  6
check my arithmetic
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