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