Let S be the statement
∃M > 0, ∀x ∈ R, (x(4 − x) < M )
That is, there is a real number M > 0 such that for all real numbers x, x(4 − x) < M .
If S is true, prove it. If S is false, disprove it by stating and proving its negation

### Statement \( S \):
The statement is as follows:

\[
S: \exists M > 0 \text{ such that } \forall x \in \mathbb{R}, \, x(4 - x) < M
\]

In words, there exists a positive real number \( M \) such that for all real numbers \( x \), the expression \( x(4 - x) \) is less than \( M \).

### Approach:
The goal is to determine if this statement is true or false by analyzing the behavior of the function \( f(x) = x(4 - x) = 4x - x^2 \).

#### 1. **Behavior of the function**:
Let's examine the function \( f(x) = 4x - x^2 \). This is a quadratic function, which is concave down (i.e., opens downwards) because the coefficient of \( x^2 \) is negative.

#### 2. **Find the maximum of \( f(x) \)**:
Since the quadratic opens downward, it will attain a maximum value at the vertex.

The vertex of a quadratic function \( f(x) = ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). In our case:

- \( a = -1 \),
- \( b = 4 \),

So, the \( x \)-coordinate of the vertex is:

\[
x = -\frac{4}{2(-1)} = 2
\]

Substitute \( x = 2 \) into the function to find the maximum value:

\[
f(2) = 4(2) - (2)^2 = 8 - 4 = 4
\]

Thus, the maximum value of \( f(x) = x(4 - x) \) is \( 4 \), which occurs at \( x = 2 \).

#### 3. **Behavior at infinity**:
As \( x \to \infty \) or \( x \to -\infty \), the quadratic term \( -x^2 \) dominates, and \( f(x) \to -\infty \). This means that as \( x \) gets large in magnitude, \( f(x) \) becomes arbitrarily negative.

#### 4. **Conclusion**:
- The function \( f(x) \) has a maximum value of \( 4 \) at \( x = 2 \), and as \( x \to \pm\infty \), the function tends to \( -\infty \).
- Since \( f(x) < 4 \) for all \( x \in \mathbb{R} \) (it is strictly less than 4 everywhere except at \( x = 2 \), where it equals 4), we can choose \( M = 4 \).

Thus, the statement \( S \) is **true** with \( M = 4 \).

### Proof:
There exists \( M = 4 \) such that for all \( x \in \mathbb{R} \):

\[
x(4 - x) \leq 4 < M
\]

This proves the statement.

### Negation:
If the statement \( S \) were false, the negation would be:

\[
\forall M > 0, \exists x \in \mathbb{R}, \text{ such that } x(4 - x) \geq M
\]

This means that for every \( M \), no matter how large, there exists some \( x \) such that \( x(4 - x) \) is at least \( M \). This is false because the function has a maximum value of 4, and it never exceeds 4 for any \( x \in \mathbb{R} \).

### Summary:
- The original statement \( S \) is **true** with \( M = 4 \).

Would you like more details on any part of this explanation?

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Here are 5 related questions to deepen your understanding:
1. How would the result change if the quadratic had a positive leading coefficient?
2. What is the significance of the vertex in finding the maximum or minimum value of a quadratic function?
3. How does the behavior of the function at infinity affect the truth of the statement?
4. Can you generalize the approach used here to other quadratic inequalities?
5. What role does concavity play in determining the bounds of a function?

**Tip**: Quadratic functions always have either a maximum or minimum value, depending on the sign of the leading coefficient. This is a useful fact when analyzing inequalities!

Let S be the statement ∃M > 0, ∀x ∈ R, (x(4 − x) < M). That is, there is a real number M > 0 such that for all real numbers x, x(4 − x) < M. If S is true, prove it. If S is false, disprove it by stating and proving its negation.

Analyze the quadratic function f(x) = x(4 - x) and determine if there exists a maximum value M such that f(x) < M for all real numbers x. This problem involves concepts of quadratic functions, inequalities, and real analysis, suitable for undergraduate students.

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