Thread: Creating Cubic Function without integrals?

1. Creating Cubic Function without integrals?

We are about to learn integrals, but this homework does not expect an understanding of them. I'm having a hard time understanding what I am to do.

The problem asks me to find cubic function in the form of $f(x)=ax^3+cx^2+d$ that has a local maximum value of 9 at -3 and a local minimum value of 7 at 0.

So I knew that the zeros for $f'(x)$ were equal to -3 and 0. Using that I came up with $x(x+3)$. I realize though that any coefficient attached to the first x would still give a zero of 0. So how do I find that coefficient?

Taking the antiderrivitive of my function, then adding the constant 7 gave: $f(x) = \frac{1}{3}x^3+\frac{3}{2}x^2+7$. However with this, $f(-3) = 11.5$, not 9. What am I missing?

EDIT:
I wanted to add that I tried also working from the second derivative, too. Thinking that the inflection point ought to be halfway between 0 and -3, I made $f''(x) = x + \frac{3}{2}$ which makes $f'(x) = \frac{1}{2}x^2 + \frac{3}{2}x$ which still has zeros of 0 and -3. The anti derivative of that gives $f(x) = \frac{1}{6}x^3 + \frac{3}{4}x^2 +7$ after adding the constant. We're really close, but $f(-3) = 9.25$ which isn't 9.

Just wanted whoever to know that I'm giving it the old college try

2. Re: Creating Cubic Function without integrals?

$f \, '(x) = Cx(x+3)$ for some constant $C.$

3. Re: Creating Cubic Function without integrals?

Take the derivative of $f(x)=ax^3+cx^2+d$. As BobP said, that's equal to $Cx(x+3)$. That gives you a relationship between a and c, so you have only two unknowns. The equation $f(0)=7$ gives you d, and the equation $f(-3)=9$ then gives you a and c.

- Hollywood

4. Re: Creating Cubic Function without integrals?

Right, I understand that much. But besides trial and error, what's the intuitive way to find that constant?

5. Re: Creating Cubic Function without integrals?

I must be extremely thick headed. I'm sorry, but I still don't get it. The derivative of $f(x) = ax^3+cx^2+d$ is $Cx(x+3)$, but 'C' isn't the 'c' from the first equation, is it? I can't see how it would be. I tried setting it up another way where $f(-3) = a(-3)^3 + c(-3)^2 + 7 = 0$ which leaves me with $3a+c=\frac{2}{9}$ but I'm not sure I'm helping myself here. I'm getting frustrated because I feel like it should be obvious, but it isn't at all.

6. Re: Creating Cubic Function without integrals?

I don't understand why you'd need integration for this.
You function is $f(x)=ax^3+cx^2+d$

The points (0,7) and (-3,9) lie on the line
$a(0)^3+c(0)^2+d=7$
$a(-3)^3+c(-3)^2+d=9$

There is a critical point at -3
$f'(x)=3x^3+2cx$
$3(-3)^3+2c(-3)=0$

Solve the simultaneous equations for a, c and d

7. Re: Creating Cubic Function without integrals?

Shakarri has the right approach, and the 3 equations you need to solve 3 variables, but messed up a bit on the 3rd equation (derivative):

$f'(x) = 3ax^2 + 2cx = 0$

with critical point at -3

$f'(x) = 0 = 3a(-3)^2 + 2c(-3)$

Should be the correct 3rd equation.

8. Re: Creating Cubic Function without integrals?

Thanks everyone. I got it figured out thanks to your help. I was frustrated because everyone kept telling me what I already knew and had said in post 1, that is, that $f'(x) = Cx(x+3)$. I just didn't realize that I needed to take the equation $f'(x) = Cx^2+3Cx$ to go to $f(x) = \frac{C}{3}x^3 + \frac{3C}{2}x^2+7$ I set that equal to 9, solved for C which was $\frac{4}{9}$ meaning $a=\frac{4}{9}*\frac{1}{3} = \frac{4}{27}$ and $c=\frac{4}{9}*\frac{3}{2} = \frac{2}{3}$. Thanks again.