quadratic equations intersection point is minimum instead of roots!

[Gevni]

I have 2 quadratic functions and I am interested in their root in the specific range. I use quadratic equation to get their roots and what I find that if their any real solution exist for both or any of the function that lie in it designated specific range, then the roots are maximum or minimum to the intersection point of range.

Let say here the intersection point is 5:

f(g) is for range [0<n<=5]
and
f(x) is for range [5<=n<10]

for f(g) real root using quadratic equation is 4.3 that lies within its range and results in equation =0 however, the minimum value of the first derivative I got is n=5 instead of n=4.3. And it is always the case and vice versa for f(x). How do I prove that intersection point in the range is always be the minimum solution?

 

[JeffM]

Sadly, that post is confused beyond hope of an answer. Let's start by clearing away confusion.

f(x) and f(g) do not represent different functions. f(x) and g(x) represent different functions on the same argument. f(g(x)) represents a third function called a composite function built from g(x) and f(x). (f(x) - g(x) = h(x) represents a third function, and where h(x) = 0 we have f(x) = g(x). What in the world are you actually talking about?

I am GUESSING that what you mean is

[MATH]f(x) \text { for } 0 \le x \le 5 \text { and } g(x) \text { for } 0 \le x \le 5.[/MATH]
If that guess is wrong, please correct it using proper notation.

In particular, do you mean

[MATH]\text {GIVEN } 0 \le x \le 5 \implies 0 \le f(x) \le 5 \text { and } f(x) \text { is continuous,}\\ \text {and } 0 \le x \le 5 \implies 0 \le g(x) \le 5 \text { and } g(x) \text { is continuous,}\\ \text {what roots if any does } f(g(x)) \text { have in } [0,\ 5]?[/MATH]Second what do quadratic equations have to do with this? You have not specified that f(x) or g(x) are differentiable, let alone polynomials.

 

[Steven G]

Your first two lines are not very clear. Can you please try again?

You can NOT say that it always works until you proof it. Up to know you think it works.

 

[Gevni]

Sorry about the confusion! Let me re-write my problem:

[MATH] {GIVEN } \\ 0 < x \le 5 \implies 0 < f(x) \le 5 \text { and } f(x) \text { is continuous,}\\ and \\ 5 \le x < 10 \implies 5 \le g(x) < 10 \text { and } g(x) \text { is continuous,}\\ [/MATH]I am only interested in real roots that lies in range specified for both individual functions. These two functions are quadratic equations that I solved for "x" using quadratic formula. However, the cost of any of function is less not with the roots that I got from quadratic formula but instead the intersection point that is 5 in above case. How to prove it?

 

[Steven G]

No, you can't conclude from 0<x<5 that 0<f(x)<5 AND that f(x) is continuous. This is generally not true. Please try again.

 

[JeffM]

OK That is a lot clearer, but we still do not know for what function you are trying to find the roots.

Are you asking whether the roots of f(x) or g(x) must be extrema? The general answer is no, but they may be.

Are you asking whether the root of h(x) = f(x) - g(x) must be an extrema? The general answer is no, but it may be. Why don't you tell us the function whose roots and extrema you are trying to find.

 

[Gevni]

I am trying to find the roots of both function individually and if roots are real and in range then the extrema is always on intersection point not on the roots that I got from quadratic equation.

 

[JeffM]

I am trying to find the roots of both function individually and if roots are real and in range then the extrema is always on intersection point not on the roots that I got from quadratic equation.

The extremum of WHAT?

 

[Gevni]

T

The extremum of WHAT?

The extremum of a function f(x) or g(x).

 

[Gevni]

No, you can't conclude from 0<x<5 that 0<f(x)<5 AND that f(x) is continuous. This is generally not true. Please try again.

I am only interested in the case if [MATH] 0<f(x)[U]<[/U]5 [/MATH] or [MATH] 5[U]< [/U]g(x)<10 [/MATH] case if it is beyond then we will not consider it as a solution. My question is if it true then the extremum of function is not the roots that I got from quadratic formula instead it is always the intersection point that is 5 in this case.

 

[JeffM]

If j(x) is a quadratic, it will have a root that is an extremum if and only if

[MATH]j(x) = ax^2 + bx + \dfrac{b^2}{4a}.[/MATH]
Proof

[MATH]j\left ( - \dfrac{b}{2a} \right ) = a * \left ( - \dfrac{b}{2a} \right )^2 + b * \left ( - \dfrac{b}{2a} \right ) + \dfrac{b^2}{4a} =\\ a * \dfrac{b^2}{4a^2} - \dfrac{b^2}{4a} + \dfrac{b^2}{4a} = \dfrac{2b^2}{4a} - \dfrac{b^2}{2a} = 0.[/MATH][MATH]j'(x) = 2ax + b \implies\\ j'\left ( - \dfrac{b}{2a} \right ) = \cancel {2a} * \left ( - \dfrac{b}{\cancel {2a}} \right ) + b = - b + b = 0.[/MATH]

 

[JeffM]

You keep talking about roots and extrema and intersections without being clear, until pressed, as to what function or functions you are talking about. It is not in general true that the extremum of f(x) or g(x) will occur at the roots of f(x) or g(x) or at the intersection of the two functions. Now of course it may happen in special cases, but it is generally false. And of course you do not tell us what the equations are, which makes it IMPOSSIBLE to figure out whether you do have such a special case. Are you deliberately trying to make it hard for us to answer your question?

 

[Steven G]

If j(x) is a quadratic, it will have a root that is an extremum if and only if

[MATH]j(x) = ax^2 + bx + \dfrac{b^2}{4a}.[/MATH]
Proof

[MATH]j\left ( - \dfrac{b}{2a} \right ) = a * \left ( - \dfrac{b}{2a} \right )^2 + b * \left ( - \dfrac{b}{2a} \right ) + \dfrac{b^2}{4a} =\\ a * \dfrac{b^2}{4a^2} - \dfrac{b^2}{4a} + \dfrac{b^2}{4a} = \dfrac{2b^2}{4a} - \dfrac{b^2}{2a} = 0.[/MATH][MATH]j'(x) = 2ax + b \implies\\ j'\left ( - \dfrac{b}{2a} \right ) = \cancel {2a} * \left ( - \dfrac{b}{\cancel {2a}} \right ) + b = - b + b = 0.[/MATH]

I think that you should recheck your calculations.

 

[JeffM]

I think that you should recheck your calculations.

Can you give me a hint where you think I have gone astray?

 

[Steven G]

Can you give me a hint where you think I have gone astray?

Now that you fixed your errors I can't.

 

[JeffM]

Now that you fixed your errors I can't.

Ahh the problem with messages crossing. I am now trying to determine the general condition for the intersection of two quadratics being the extremum of one of them.

 

[Gevni]

You keep talking about roots and extrema and intersections without being clear, until pressed, as to what function or functions you are talking about. It is not in general true that the extremum of f(x) or g(x) will occur at the roots of f(x) or g(x) or at the intersection of the two functions. Now of course it may happen in special cases, but it is generally false. And of course you do not tell us what the equations are, which makes it IMPOSSIBLE to figure out whether you do have such a special case. Are you deliberately trying to make it hard for us to answer your question?

Sorry JeffM, If you feel like that. Actually I am not a mathematician, but trying to use math in real life problem and my math base is very week I admit, so I think I won't explain properly my point. I will try to re-write it again with real functions

 

[JeffM]

Good idea Gevni. It always helps to start with more information than necessary rather then less. In the meantime, I am working on the general conditions for an extremum of one quadratic to occur at its intersection with another quadratic.

 

[Gevni]

So my functions are:
[MATH] f(x) =(A-B)X^2 +2BnX-Bn^2 \space \space \space range [0 <x <= t] \\ g(x)=(C+D)X^2 -2CnX+Cn^2 \space \space \space range [ t<=x <n] [/MATH]Here , A,B, C,D,n,t are known. I solved these two functions using quadratic formula and only interested in solution of f(x) or g(x) if any or both have real and within specific range roots. After I consider real and within range roots and intersection point (t) as candidate value of x that will minimize the cost of my formula f(x) or g(x). When I compare the cost of all candidates, intersection point will always give me the lowest cost other then any minimum point in roots . I have to prove it in general. I hope now it is clear. So basically, I am using quadratic equation to solve the optimization problem. The lowest value of function will be the optimal one and my result shows, that I got optimal results when I use intersection point as value of x instead of any real and within range root.

 

[JeffM]

So my functions are:
[MATH] f(x) =(A-B)X^2 +2BnX-Bn^2 \space \space \space range [0 <x <= t] \\ g(x)=(C+D)X^2 -2CnX+Cn^2 \space \space \space range [ t<=x <n] [/MATH]Here , A,B, C,D,n,t are known. I solved these two functions using quadratic formula and only interested in solution of f(x) or g(x) if any or both have real and within specific range roots. After I consider real and within range roots and intersection point (t) as candidate value of x that will minimize the cost of my formula f(x) or g(x). When I compare the cost of all candidates, intersection point will always give me the lowest cost other then any minimum point in root. I have to prove it in general. I hope now it is clear.

Now we are getting somewhere.

What do you mean, you solved these functions? You solve equations, not functions. In particular, you use the quadratic formula to solve equations of the form quadratic expression = 0. This is also frequently formulated that you use the quadratic formula to find the roots of a quadratic function because the definition of a root is f(root) = 0. The root tells you where the function has a value of zero. So I am a little dubious about the validity of a cost function in the real world having a root in a feasible range because it is seldom practical to reduce a material cost to zero.

With me to here? I am going to assume that a - b > 0 and c + d > 0. That means we do not have to bother with second or higher order derivatives.

By all means, solve for the roots of f(x) and g(x) using the quadratic formula. In the unlikely event that you find any roots in a feasible range, you have your answer: you can bring the cost down to zero. If that does not work, find the roots of the first derivatives of f(x) and g(x). (Do you know how to do that?) If only one of those roots is in a feasible range, you have your answer. If both are in a feasible range, evaluate f(x) at the root of its first derivative and evaluate g(x) at the root of its first derivative. Whichever evaluation gives you a lower cost gives you your answer. If neither first derivative has a root in a feasible range, evaluate f(0), f(t), g(t), and (g(n). Whichever evaluation gives you the lowest cost gives you your answer.

As far as I can see, the intersection of f(x) and g(x) is not helpful in any way.