Math Problem Statement

2/x+1 > x-1/x+1

Solution

To solve the inequality

2x+1>x1x+1\frac{2}{x+1} > x - \frac{1}{x+1}

let's follow these steps:

Step 1: Combine like terms

We can rewrite the inequality as:

2x+1>x(x+1)1x+1\frac{2}{x+1} > \frac{x(x+1) - 1}{x+1}

Since the right-hand side already has a common denominator x+1x+1, we can subtract the fractions:

2x+1x(x+1)1x+1>0\frac{2}{x+1} - \frac{x(x+1) - 1}{x+1} > 0

Step 2: Simplify the expression

Combine the fractions:

2(x(x+1)1)x+1>0\frac{2 - \left(x(x+1) - 1\right)}{x+1} > 0

Now, expand and simplify the numerator:

2(x2+x1)=2x2x+1=3x2x2 - (x^2 + x - 1) = 2 - x^2 - x + 1 = 3 - x^2 - x

So the inequality becomes:

3x2xx+1>0\frac{3 - x^2 - x}{x+1} > 0

Step 3: Factor the numerator

The expression in the numerator can be factored:

(x2+x3)x+1>0\frac{-(x^2 + x - 3)}{x+1} > 0

We can factor the quadratic x2+x3x^2 + x - 3 as:

x2+x3=(x1)(x+3)x^2 + x - 3 = (x - 1)(x + 3)

So, the inequality is:

(x1)(x+3)x+1>0\frac{-(x - 1)(x + 3)}{x+1} > 0

Step 4: Analyze the sign changes

We need to determine the intervals where the expression is positive. The critical points occur where the numerator or denominator is zero:

  • x1=0x=1x - 1 = 0 \Rightarrow x = 1
  • x+3=0x=3x + 3 = 0 \Rightarrow x = -3
  • x+1=0x=1x + 1 = 0 \Rightarrow x = -1

We will now test the intervals determined by these points:

  1. x<3x < -3
  2. 3<x<1-3 < x < -1

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Math Problem Analysis

Mathematical Concepts

Inequalities
Rational Functions
Quadratic Equations

Formulas

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Theorems

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Suitable Grade Level

Grades 10-12