Math  /  Algebra

Questionx2+2xx+1<x+3\frac{x^{2}+2 x}{x+1}<x+3

Studdy Solution

STEP 1

1. The inequality involves a rational expression on the left-hand side.
2. The inequality can be solved by finding critical points and testing intervals.
3. The solution will involve considering the domain of the rational expression.

STEP 2

1. Simplify the inequality.
2. Find critical points by solving the equation and undefined points.
3. Test intervals determined by the critical points.
4. Determine the solution set.

STEP 3

First, simplify the inequality. Start by subtracting x+3 x + 3 from both sides:
x2+2xx+1(x+3)<0\frac{x^2 + 2x}{x+1} - (x + 3) < 0
Combine the terms over a common denominator:
x2+2x(x2+3x+3x+3)x+1<0\frac{x^2 + 2x - (x^2 + 3x + 3x + 3)}{x+1} < 0
Simplify the numerator:
x2+2xx26x3x+1<0\frac{x^2 + 2x - x^2 - 6x - 3}{x+1} < 0
This simplifies to:
4x3x+1<0\frac{-4x - 3}{x+1} < 0

STEP 4

Find critical points by setting the numerator and denominator equal to zero:
1. Numerator: 4x3=0-4x - 3 = 0 $ -4x = 3 \quad \Rightarrow \quad x = -\frac{3}{4} \]
2. Denominator: x+1=0x + 1 = 0 $ x = -1 \]
The critical points are x=34x = -\frac{3}{4} and x=1x = -1.

STEP 5

Test intervals determined by the critical points: (,1) (-\infty, -1) , (1,34)(-1, -\frac{3}{4}), and (34,)(- \frac{3}{4}, \infty).
1. Interval (,1)(- \infty, -1): Choose x=2x = -2: $ \frac{-4(-2) - 3}{-2 + 1} = \frac{8 - 3}{-1} = -5 < 0 \quad \text{(True)} \]
2. Interval (1,34)(-1, -\frac{3}{4}): Choose x=0.9x = -0.9: $ \frac{-4(-0.9) - 3}{-0.9 + 1} = \frac{3.6 - 3}{0.1} = 6 > 0 \quad \text{(False)} \]
3. Interval (34,)(- \frac{3}{4}, \infty): Choose x=0x = 0: $ \frac{-4(0) - 3}{0 + 1} = \frac{-3}{1} = -3 < 0 \quad \text{(True)} \]

STEP 6

Determine the solution set based on the intervals where the inequality holds true:
The solution set is (,1)(34,) (-\infty, -1) \cup (-\frac{3}{4}, \infty) .
The solution to the inequality is:
(,1)(34,)\boxed{(-\infty, -1) \cup (-\frac{3}{4}, \infty)}

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