How Do You Determine the Coefficients of a Cubic Function with Given Extremes?

[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

 

[arildno]

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

 

[Jeff Ford]

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

 

[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 separately as unknown variables. Just find the right cubic and then you just can read them off.

 

[arildno]

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

 

[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.

 

[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

 

[benorin]

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

 

[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.

 

[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.

 

[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

 

[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.

 

[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?