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

Questions
1

The question is:

2025-07-27_190919.png

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
2

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]]
3

Good. Address the problem with universal and existential quantifiers, along with two additional existence constraints.

4

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]
5

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.

6

When 0<=m<1 the minimum value is larger than 2.

7

Which would appear to make the point m=1 irrelevant for purposes of the question posed

8

Perhaps you read “a minimum value” as meaning “a lower bound”?

9

Rereading the question, I will maintain that the given function, defined on [0,m], has satisfies 2<=f<=3 for m=2. I do not see where 2 separate values are requested for m, and I see no restriction that f take a min at an endpoint. My main point here is that the question asks to find the value m.

10

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.