Math Problem Statement

Analyzing the function f(x) = (x^4 - 2x^2 + 3) / (-x^4 + 2x^2), determine its domain, symmetry, specific expression forms, boundedness, limits, and monotonicity on specific intervals.

Solution

Let’s analyze and solve the given mathematical exercise step by step.

The function f(x)f(x) is defined as: f(x)=x42x2+3x4+2x2.f(x) = \frac{x^4 - 2x^2 + 3}{-x^4 + 2x^2}.

1. (a) Déterminer DD, l'ensemble de définition de ff.

The function f(x)f(x) is defined wherever the denominator is non-zero. The denominator is: x4+2x2=x2(x2+2).-x^4 + 2x^2 = x^2(-x^2 + 2). Setting the denominator to zero: x2(x2+2)=0.x^2(-x^2 + 2) = 0. This occurs when x=0x = 0 or x2=2x^2 = 2. Therefore, x=±2x = \pm \sqrt{2}.

The domain DD of ff is: D=R{0,±2}.D = \mathbb{R} \setminus \{ 0, \pm\sqrt{2} \}.


1. (b) Montrer que ff est paire.

A function f(x)f(x) is even if f(x)=f(x)f(-x) = f(x) for all xx in the domain DD.

Given f(x)=x42x2+3x4+2x2f(x) = \frac{x^4 - 2x^2 + 3}{-x^4 + 2x^2}, note that all powers of xx in both the numerator and denominator are even, so substituting x-x results in: f(x)=(x)42(x)2+3(x)4+2(x)2=x42x2+3x4+2x2=f(x).f(-x) = \frac{(-x)^4 - 2(-x)^2 + 3}{-(-x)^4 + 2(-x)^2} = \frac{x^4 - 2x^2 + 3}{-x^4 + 2x^2} = f(x). Thus, f(x)f(x) is even.


2. (a) Montrer que pour tout xDx \in D, f(x)=3x2(2x2)1f(x) = \frac{3}{x^2(2 - x^2)} - 1.

Starting with f(x)=x42x2+3x4+2x2f(x) = \frac{x^4 - 2x^2 + 3}{-x^4 + 2x^2}, factor the numerator and denominator: Numerator: x42x2+3.\text{Numerator: } x^4 - 2x^2 + 3. Denominator: x4+2x2=x2(x2+2).\text{Denominator: } -x^4 + 2x^2 = x^2(-x^2 + 2). Rewriting f(x)f(x): f(x)=x42x2+3x2(x2+2)=x2(x22)+3x2(2x2).f(x) = \frac{x^4 - 2x^2 + 3}{x^2(-x^2 + 2)} = \frac{x^2(x^2 - 2) + 3}{x^2(2 - x^2)}. Split the fraction: f(x)=3x2(2x2)1.f(x) = \frac{3}{x^2(2 - x^2)} - 1.


2. (b) Montrer que ff est minorée sur ]0,2[]0, \sqrt{2}[.

To show ff is bounded below, analyze the expression: f(x)=3x2(2x2)1.f(x) = \frac{3}{x^2(2 - x^2)} - 1. For x]0,2[x \in ]0, \sqrt{2}[, note that x2(2x2)>0x^2(2 - x^2) > 0, so 3x2(2x2)>0\frac{3}{x^2(2 - x^2)} > 0. Therefore: f(x)=3x2(2x2)1>1.f(x) = \frac{3}{x^2(2 - x^2)} - 1 > -1. Thus, f(x)f(x) is bounded below by 1-1 on ]0,2[]0, \sqrt{2}[.


3. (a) Calculer limx+f(x)\lim_{x \to +\infty} f(x).

As x+x \to +\infty, the dominant term in the numerator and denominator is x4x^4. Thus: f(x)=x42x2+3x4+2x2x4x4=1.f(x) = \frac{x^4 - 2x^2 + 3}{-x^4 + 2x^2} \sim \frac{x^4}{-x^4} = -1. Therefore: limx+f(x)=1.\lim_{x \to +\infty} f(x) = -1.


3. (b) Montrer que ff est croissante sur ]2,+[]\sqrt{2}, +\infty[.

To analyze monotonicity, compute f(x)f'(x). Rewrite f(x)f(x): f(x)=3x2(2x2)1.f(x) = \frac{3}{x^2(2 - x^2)} - 1. Differentiate f(x)f(x): f(x)=ddx(3x2(2x2)).f'(x) = -\frac{d}{dx} \left( \frac{3}{x^2(2 - x^2)} \right). Using the quotient rule, compute f(x)f'(x) explicitly. Then verify that f(x)>0f'(x) > 0 for x>2x > \sqrt{2}, which shows f(x)f(x) is increasing on ]2,+[]\sqrt{2}, +\infty[.

Would you like to see detailed steps for f(x)f'(x) or any specific section above? Here are follow-up questions for further exploration:


Follow-Up Questions:

  1. How is the symmetry of a function verified in general?
  2. Could you explain in detail how to compute f(x)f'(x) for this function?
  3. How do we ensure f(x)f(x) is bounded below on specific intervals?
  4. What techniques are used to evaluate limits of rational functions as xx \to \infty?
  5. Why is the domain of the function restricted to exclude specific points?

Tip:

Always simplify the numerator and denominator fully before analyzing properties like limits, symmetry, or monotonicity.

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Math Problem Analysis

Mathematical Concepts

Functions and their properties
Rational functions
Limits and continuity
Symmetry in functions
Derivatives and monotonicity

Formulas

f(x) = (x^4 - 2x^2 + 3) / (-x^4 + 2x^2)
Domain exclusion: Denominator ≠ 0
Limit: lim (x → ∞) of rational functions
Derivative: Quotient rule

Theorems

Even function property: f(-x) = f(x)
Monotonicity: f'(x) > 0 or f'(x) < 0 in an interval
Limit evaluation for dominant terms in rational functions

Suitable Grade Level

Undergraduate (Calculus I or II)