Constraints problem

[Deleted member 4993]

That is my mistake, the constraints I need to find are of a, b, c, d, and e

OP has "a, b & c" as unknowns. Where did "d & e" come from?

 

[hajfajv]

Actually, we haven't ever been given the actual wording of a problem; this may be part of a larger problem, and not necessarily well understood. Certainly the wording we have contributes to the impression that the goal would be to have two critical points, for which we'd expect an inequality constraint.

Solving for actual values clarifies the "constraint"! (As I read it, the lack of a constraint on c would also be part of the answer!)

@hajfajv, can you show us the entire problem you are working on (both in the original language and a translation, if it is not in English)? That will help us know what you really need.

Here is a picture taken from my book

 

[hajfajv]

OP has "a, b & c" as unknowns. Where did "d & e" come from?

oh I wrote them from the problem above in my book, whoops

 

[hajfajv]

I'll check back later my tablet needs power

 

[JeffM]

I don't understand line three of the equations, where does the greater then or equalsign come from?

So we are talking about the generic cubic function

[math]f(x) = px^3 + qx^2 + sx + t, \text { where } p, \ q, \ s, \text { and } t \in \mathbb R \text { and } p \ne 0.[/math]
(Don’t ask me why I skipped r). The first derivative is a quadratic function, and the second derivative is a linear function. If a cubic function has local extrema, its first derivative will equal zero there. But that derivative is a quadratic, and thus may have two distinct zeroes, a single value at which it is zero, or no zero. How do we determine which of those three cases is true? We look at the determinant.

[math]\alpha x^2 + \beta x + \gamma = 0 \implies\\ \text {two distinct solutions} \iff \beta^2 - 4 \alpha \gamma > 0 \iff \beta^2 > 4 \alpha \gamma;\\ \text {one solution} \iff \beta^2 - 4 \alpha \gamma= 0 \iff \beta^2 = 4\alpha \gamma;\\ \text {and no real solution} \iff \beta^2 - 4 \alpha \gamma < 0 \iff \beta^2 < 4 \alpha \gamma.[/math]
Remember that from first year of algebra: the quadratic formula?

OK. Our derivative is [imath]3px^2 + 2qx + s[/imath] Going to the determinant of that quadratic, [imath](2q)^2 = 4q^2[/imath] and [imath]4 * 3p * s = 12ps[/imath]. The derivative is nowhere zero if [imath]4q^2 < 12ps.[/imath] But the derivative does have at least one zero if [imath]4q^2 \ge 12ps.[/imath]

Now if you think about the general cubic, the graph will generally be rising or falling, but it can have a local maximum, in which case it must also have a local minimum. In other words, the cubic is interesting if the derivative has two zeroes. What if it has only one zero? If you follow that through, the second derivative also has a zero at the same value. There is no extremum there, just an inflection point.

 

[Dr.Peterson]

oh I can just subtract 2a, I divided like I was just getting rid of [imath]a[/imath], thanks this should clear this up

I assume you realize that dividing as you did doesn't "get rid of a"; it results in a non-equivalent equation! The correct way to "get rid of a" is to subtract, as you now realize, because you need to "undo" an addition.

 

[hajfajv]

Now I found the constraints for [imath]a[/imath] and [imath]b[/imath] to be [imath]-6[/imath] and [imath]9[/imath] respectively.

 

[Dr.Peterson]

Now I found the constraints for [imath]a[/imath] and [imath]b[/imath] to be [imath]-6[/imath] and [imath]9[/imath] respectively.
View attachment 31757

Good. So the answer to the question "what requirements must be imposed on a, b, and c" is that [imath]a=-6, b=9,[/imath] and c can be any real number. Correct?

And, of course, we can check it. We have [imath]f(x) = x^3-6x^2+9x+c[/imath]:



There I took [imath]c=-3[/imath]. For other values, the graph shifts up and down.

 

[hajfajv]

Good. So the answer to the question "what requirements must be imposed on a, b, and c" is that [imath]a=-6, b=9,[/imath] and c can be any real number. Correct?

And, of course, we can check it. We have [imath]f(x) = x^3-6x^2+9x+c[/imath]:

View attachment 31758

There I took [imath]c=-3[/imath]. For other values, the graph shifts up and down.

Sure c can be whatever. Also I'm not used to graphing, for this class there are no calculators of any kind.

 

[Dr.Peterson]

I'm not used to graphing, for this class there are no calculators of any kind.

The graph is not for doing the work, but for showing that you got it right (and maybe make it clear after the fact what the problem is about). Desmos.com, which I used, is very helpful outside of class as a learning aid (though not as a crutch).

 

[Steven G]

Just for the record, you were lucky that when you equated both equations they each had a y on both sides (not say 2y on one side and 7y on the other side) and they easily cancelled out.