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Interval boundary m when a quadratic has min 2 and max 3

Jim Clinton
Posted 1 month ago

The question is: enter image description here

f[x_] := x^2 - 2 x + 3
Reduce[ForAll[x, 0 <= x <= m, 2 <= f[x] <= 3]]

The code above produces an incorrect result.

m <= 2

The accurate result is given below. How should the code be modified to match it?

1 <= m <= 2
POSTED BY: Jim Clinton
9 Replies
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I cannot see how 1 figures into this as a lower bound on m. Based on the wording of the question and the math. Only 2 seems relevant.

POSTED BY: Daniel Lichtblau

When 0<=m<1 the minimum value is larger than 2.
POSTED BY: Gianluca Gorni

Which would appear to make the point m=1 irrelevant for purposes of the question posed
POSTED BY: Daniel Lichtblau

Perhaps you read "a minimum value" as meaning "a lower bound"?
POSTED BY: Gianluca Gorni
POSTED BY: Daniel Lichtblau

In my personal reading, the condition of the question is that 2<=f<=3 AND that 2 and 3 are attained somewhere. The final question "the number m is ___" is poorly worded, because it seems to ask for a single solution. Also, "a minimum value" annoys me, because the minimum value is unique, if it exists.

However, I am not a native English speaker.
POSTED BY: Gianluca Gorni

You could translate "minimum value" as MinValue[...] and "maximum value" as MaxValue[...]. Then the problem can be formulated as follows:

Reduce[{
  MinValue[{x^2 - 2 x + 3, 0 <= x <= m}, x] == 2,
  MaxValue[{x^2 - 2 x + 3, 0 <= x <= m}, x] == 3},
 {m}, Reals]
POSTED BY: Michael Rogers

I would do it this way:

Reduce[ForAll[x, 0 <= x <= m, 2 <= f[x] <= 3] && 
  Exists[x, 0 <= x <= m, 2 == f[x]] && 
  Exists[x, 0 <= x <= m, f[x] == 3]]
POSTED BY: Gianluca Gorni
Jim Clinton
Jim Clinton
Posted 26 days ago

Good. Address the problem with universal and existential quantifiers, along with two additional existence constraints.
POSTED BY: Jim Clinton
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