Posted by **Danielle** on Thursday, February 26, 2009 at 10:42pm.

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=3-27b.

- Calculus -
**drwls**, Thursday, February 26, 2009 at 11:02pm
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

- Calculus -
**Danielle**, Thursday, February 26, 2009 at 11:20pm
I understand everything up until the last part with the using equations. Where did 8a and 4b come from?

- Calculus -
**Reiny**, Thursday, February 26, 2009 at 11:41pm
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 x-axis, there has to be a double root at x=2

so we have (x-2)(x-2) as factors

since it is a cubic, there can be only one other linear factor

so let the function be

y = a(x-b)(x-2)^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(x-b)(x-2)^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)(x-2)^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.

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