# Homework Help: Find a cubic function

1. Mar 26, 2006

### Jeff Ford

Find a cubic function $$f(x) = ax^3 - bx^2 + cx - d$$ that has a local maximum value of 40 at x = 1 and a local minimum valud of -68 at x = 4

Since x = 1 and x = 4 are the max and min, respectively, then f'(x) must equal 0 at x = 1 and x = 4. Therefore $$f'(x) = (x-1)(x-4) = x^2 - 5x +4$$

Using the original function $$f'(x) = 3ax^2 - 2bx +c = x^2 - 5x + 4$$

Setting the terms equal to each other we get
$$3ax^2 = x^2 \Rightarrow a = \frac{1}{3}$$
$$-2bx = -5x \Rightarrow b = \frac{5}{2}$$
$$c = 4$$

Substituting these values back into the original equation I get

$$f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 4 - d$$

When I solve for d using x = 1 and f(x) = 40 I get

$$d = \frac{1}{3} - \frac{5}{2} - 36 = \frac{-229}{6}$$

However, when I solve for d using x = 4 and f(x) = -68 I get

$$d = \frac{4^3}{3} - \frac{5(4^2)}{2} + 84 = \frac{226}{3}$$

Any ideas where I went wrong, or if I was down the wrong track to start with?

Thanks
Jeff

2. Mar 26, 2006

### arildno

Eeh, from what I can see, we must have:
$$f'(x)=K(x-1)(x-4)$$, where K is some arbitrary constant.

3. Mar 26, 2006

### Jeff Ford

So if that was brought into the equation, would I be solving for a, b, c, & d in terms of K?

4. Mar 26, 2006

### StatusX

No, you need to solve for a value of K and a value for the integration constant which give you the desired values at x=1 and 4. And don't worry about solving a,b,c,d seperately as unknown variables. Just find the right cubic and then you just can read them off.

5. Mar 26, 2006

### arildno

Well, there was nothing inherently wrong about Jeff's idea. A bit cumbersome, that's all.

6. Mar 26, 2006

### Jeff Ford

Sorry, not familiar with the idea of an itegration constant. We just finished sections of optimization problems and Newton's method. The next section is antiderivatives. So I have to solve this without any concept of integration.

7. Mar 27, 2006

### Jeff Ford

Could someone direct me towards information on finding the integration constant, if that is indeed the most direct way to solve this problem?

Thanks
Jeff

8. Mar 27, 2006

### benorin

Scratch the whole integration constant idea gig: no integration for you.

9. Mar 27, 2006

### HallsofIvy

f'= k(x-1)(x- 4)= kx^2- 5kx+ 4k= 3ax^2+ bx+ c so
a= k/3, b= -5k and c= 4k. Evaluate f(x)= (k/3)x^3- 5kx^2+ 4kx+ d at 1 and 4 to get two equations for k and d.

10. Mar 27, 2006

### benorin

We have 4 unknowns, hence we need 4 equations to solve: they are f(1)=40, f(4)=-68, f'(1)=0, and f'(4)=0.

11. Mar 27, 2006

### J77

Where you went wrong in finding d in your OP was substituting the values at f'(x)=0 into f(x)...

You may also want to use: f''(1)<0, f''(4)>0

12. Mar 27, 2006

### Jeff Ford

Thanks to all. I used K(x-1)(x-4) = 0, solved for a, b, and c, then went back to the original equation and used x = 1 and x = 4 to get two equations for d in terms of K. I set them equal, solved for K, and when I plugged it back in I ended up with values for a, b, c, and d that satisfied all conditions.

13. May 15, 2007

### joyjoy61

Can anyone explain this further?

Can anyone explain this further? I have a similar problem, and I'm having trouble following what the correct answer Jeff figured out in the end was.
How does one make an equation for d in terms of K, for instance?