Math Problem Statement

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

Solution

Statement SS:

The statement is as follows:

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

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

Approach:

The goal is to determine if this statement is true or false by analyzing the behavior of the function f(x)=x(4x)=4xx2f(x) = x(4 - x) = 4x - x^2.

1. Behavior of the function:

Let's examine the function f(x)=4xx2f(x) = 4x - x^2. This is a quadratic function, which is concave down (i.e., opens downwards) because the coefficient of x2x^2 is negative.

2. Find the maximum of f(x)f(x):

Since the quadratic opens downward, it will attain a maximum value at the vertex.

The vertex of a quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c occurs at x=b2ax = -\frac{b}{2a}. In our case:

  • a=1a = -1,
  • b=4b = 4,

So, the xx-coordinate of the vertex is:

x=42(1)=2x = -\frac{4}{2(-1)} = 2

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

f(2)=4(2)(2)2=84=4f(2) = 4(2) - (2)^2 = 8 - 4 = 4

Thus, the maximum value of f(x)=x(4x)f(x) = x(4 - x) is 44, which occurs at x=2x = 2.

3. Behavior at infinity:

As xx \to \infty or xx \to -\infty, the quadratic term x2-x^2 dominates, and f(x)f(x) \to -\infty. This means that as xx gets large in magnitude, f(x)f(x) becomes arbitrarily negative.

4. Conclusion:

  • The function f(x)f(x) has a maximum value of 44 at x=2x = 2, and as x±x \to \pm\infty, the function tends to -\infty.
  • Since f(x)<4f(x) < 4 for all xRx \in \mathbb{R} (it is strictly less than 4 everywhere except at x=2x = 2, where it equals 4), we can choose M=4M = 4.

Thus, the statement SS is true with M=4M = 4.

Proof:

There exists M=4M = 4 such that for all xRx \in \mathbb{R}:

x(4x)4<Mx(4 - x) \leq 4 < M

This proves the statement.

Negation:

If the statement SS were false, the negation would be:

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

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

Summary:

  • The original statement SS is true with M=4M = 4.

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


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!

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Math Problem Analysis

Mathematical Concepts

Inequalities
Quadratic Functions
Real Analysis

Formulas

f(x) = x(4 - x) = 4x - x^2
Vertex formula: x = -b / 2a for quadratic functions ax^2 + bx + c

Theorems

Quadratic Maximum Theorem: For a quadratic function with negative leading coefficient, the maximum value occurs at the vertex.

Suitable Grade Level

Undergraduate