Find a cubic function...


by Jeff Ford
Tags: cubic, function
Jeff Ford
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#1
Mar26-06, 11:08 AM
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Find a cubic function [tex] f(x) = ax^3 - bx^2 + cx - d [/tex] 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 [tex] f'(x) = (x-1)(x-4) = x^2 - 5x +4 [/tex]

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

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

Substituting these values back into the original equation I get

[tex] f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 4 - d [/tex]

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

[tex] d = \frac{1}{3} - \frac{5}{2} - 36 = \frac{-229}{6} [/tex]

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

[tex] d = \frac{4^3}{3} - \frac{5(4^2)}{2} + 84 = \frac{226}{3} [/tex]

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

Thanks
Jeff
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arildno
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#2
Mar26-06, 11:22 AM
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Eeh, from what I can see, we must have:
[tex]f'(x)=K(x-1)(x-4)[/tex], where K is some arbitrary constant.
Jeff Ford
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#3
Mar26-06, 12:38 PM
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So if that was brought into the equation, would I be solving for a, b, c, & d in terms of K?

StatusX
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#4
Mar26-06, 01:14 PM
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Find a cubic function...


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.
arildno
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#5
Mar26-06, 01:21 PM
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Well, there was nothing inherently wrong about Jeff's idea. A bit cumbersome, that's all.
Jeff Ford
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#6
Mar26-06, 04:43 PM
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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.
Jeff Ford
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#7
Mar27-06, 07:27 AM
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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
benorin
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#8
Mar27-06, 07:29 AM
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Scratch the whole integration constant idea gig: no integration for you.
HallsofIvy
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#9
Mar27-06, 07:46 AM
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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.
benorin
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#10
Mar27-06, 07:51 AM
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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.
J77
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#11
Mar27-06, 08:15 AM
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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
Jeff Ford
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#12
Mar27-06, 09:32 AM
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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.
joyjoy61
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#13
May15-07, 06:24 PM
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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?


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