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quadratic equations intersaction point is minimum instead of roots!

gevni

New member
May 3, 2020
17
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?
 

Theia

Well-known member
Mar 30, 2016
90
in some subspace
Can you clarify your question please? I didn't understand, what curves intersect: two parabolic curves each others, or parabolic curve and x-axis, or something else? Maybe provide a numerical example about your problem? That could make your question clearer too.(up)
 

gevni

New member
May 3, 2020
17
Sorry about the confusion! Let me re-write my problem:

\(\displaystyle
{GIVEN } \\
0 < x \le 5 \implies 0 < f(x) \le 5 \text { and } f(x) \text { is continuous,}\\
and \\
5 \le x < 10 \implies 5 \le g(x) < 10 \text { and } g(x) \text { is continuous,}\\
\)

if f(x) or g(x) are not in the range then I am not interested, I am only interested in the case if roots are real and in range. 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, that is 5 in above case. How to prove it?
 
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