Calculus
posted by Danielle on .
I'm desperate.. I keep getting b=0 and I know it's wrong!!
Find a cubic function f(x) = ax^3 + bx^2 + cx + d that has a local maximum at (3,3) and a local minimum at (2,0) knowing that c=12b and d=327b.

You can reduce the number of unkowns to 2 (a and b) by rewriting the equatiosn with substitutions for c and d.
The system appears overdetermined. You could write four equations in two unknowns knowing the value of two points and the fact that f'(x) = 0 at those points. I would try to use the facts that
f'(x) = 3ax^2 + 2bx 12b = 0 at x = 3 and x=2 to solve for a and b.
27a 6b 12b = 27a 18b = 0
12a +4b 12b = 12a 8b = 0
Those last two equations are equivalent; you can only use one of them. So you need to use an equation that says f(2) = 0, for example.
8a + 4b [12b*(2)] +3 27b = 0
8a + b = 3
24a + 3b = 9
24a 16b = 0
19 b = 9
This does not lead to an integer value for b. I may have made a mistake somewhere. Check my work and thinking 
I understand everything up until the last part with the using equations. Where did 8a and 4b come from?

I too got myself all caught up in nasty fractions, so I tried a different approach.
Since (2,0) is a minimum and it touches the xaxis, there has to be a double root at x=2
so we have (x2)(x2) as factors
since it is a cubic, there can be only one other linear factor
so let the function be
y = a(xb)(x2)^2 , my b is not necessarily the same as the drwls's b)
I then found the derivative of that and subbed in the fact that if x=3, dy/dx = 0
this led to 55a + 10ab = 0
a(55 + 10b) = 0
a = 0, not possible if we want a cubic
or
b = 11/2
I then subbed in (3,3) in y = a(xb)(x2)^2 and using the fact that b=11/2 I got
a = 6/125
so my function is
f(x) = (6/125)(x+11/2)(x2)^2
both of your points (2,0) and (3,3) satisfy,
I then found the derivative of that function and set it equal to zero
that gave me x = 2, or x = 3
Q.E.D.
I will leave it up to your to expand it.
As drwls also noted, there seems to be redundant information.