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Find a cubic function... 
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#1
Mar2606, 11:08 AM

P: 159

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) = (x1)(x4) = 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 


#2
Mar2606, 11:22 AM

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P: 12,016

Eeh, from what I can see, we must have:
[tex]f'(x)=K(x1)(x4)[/tex], where K is some arbitrary constant. 


#3
Mar2606, 12:38 PM

P: 159

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



#4
Mar2606, 01:14 PM

HW Helper
P: 2,567

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.



#5
Mar2606, 01:21 PM

Sci Advisor
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P: 12,016

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



#6
Mar2606, 04:43 PM

P: 159

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
Mar2706, 07:27 AM

P: 159

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
Mar2706, 07:29 AM

HW Helper
P: 1,025

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



#9
Mar2706, 07:46 AM

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PF Gold
P: 39,553

f'= k(x1)(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
Mar2706, 07:51 AM

HW Helper
P: 1,025

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
Mar2706, 08:15 AM

P: 1,157

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
Mar2706, 09:32 AM

P: 159

Thanks to all. I used K(x1)(x4) = 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
May1507, 06:24 PM

P: 1

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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