Max and Min of a poly

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  • #1
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Main Question or Discussion Point

I need to make to functions in java that gives the maxim and minin of the Parabola polynom ax2+bx+c for an interval of two given points.

I have no Idea how to make this algorithm , could you help ?

I have come to something like this :
if (-b/(2*a)>=x1 && -b/(2*a)<=x2)
return (-b/(2*a));
 

Answers and Replies

  • #2
Hurkyl
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I have no Idea how to make this algorithm , could you help ?
Figure out the math first, then worry about how to program it.
 
  • #3
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Figure out the math first, then worry about how to program it.
I dont know the math that's why I'm asking. I dont want the java code.
 
  • #4
Hurkyl
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Well, what do you know about finding minima and maxima?

Alternatively, what do you know about the shape of the graphs of parabolas?
 
  • #5
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Well, what do you know about finding minima and maxima?

Alternatively, what do you know about the shape of the graphs of parabolas?
well the maxima should be the value of Y which is the bigger to a value of X and the minima the same.

About the shape its sinusoidal waves.
 
  • #6
crd
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well the maxima should be the value of Y which is the bigger to a value of X and the minima the same.

About the shape its sinusoidal waves.
Careful, you are mixing apples and oranges. The max(min) will be the value of Y which is bigger(smaller) than every other value of Y for some region around your max(min) value.

Not X like you said, X is the input variable that determines your Y.

You should try graphing ax^2 + bx + c for various values of a, b, and c to verify if it has "sinusoidal waves."

that also might give you some intuition into the the max(min) of a parabola
 
  • #7
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Theorem: If E ⊂ R and f: E → R, and f has a maximum or minimum at x ∈ E, then one of the following three is true:
(1) x is a boundary point of E,
(2) f'(x) = 0, or
(3) f is not differentiable at x.

In your case, f(x) = ax2 + bx + c and E is the interval [x1, x2]. Then the only possibilities are these: (1) x is one of the boundary points x1 or x2 of E, or (2) f'(x) = 2ax + b = 0, so x = -b/2a. Look at the values of f at those three points; the largest one is the maximum, and the smallest one is the minimum.
 

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