Math  /  Algebra

QuestionSolve the inequality x+8x+1<4\frac{x+8}{x+1}<-4 Give your answer in interval notation.

Studdy Solution

STEP 1

What is this asking? We need to find all the xx values that make the fraction x+8x+1\frac{x+8}{x+1} less than negative four. Watch out! Don't multiply both sides by x+1x+1 right away!
We don't know if x+1x+1 is positive or negative, so we don't know whether to flip the inequality sign.

STEP 2

1. Rewrite the inequality
2. Find the critical points
3. Test the intervals

STEP 3

Let's move everything to one side to get a zero on the other.
We'll add four to both sides of our inequality: x+8x+1+4<0 \frac{x+8}{x+1} + 4 < 0

STEP 4

Now, let's combine the terms on the left side into a single fraction.
Remember, 44 is the same as 4(x+1)x+1\frac{4(x+1)}{x+1}.
So, we have: x+8x+1+4(x+1)x+1<0 \frac{x+8}{x+1} + \frac{4(x+1)}{x+1} < 0 x+8+4(x+1)x+1<0 \frac{x+8 + 4(x+1)}{x+1} < 0 x+8+4x+4x+1<0 \frac{x+8 + 4x+4}{x+1} < 0 5x+12x+1<0 \frac{5x+12}{x+1} < 0 Much better!

STEP 5

**Critical points** are where the numerator is zero or the denominator is zero.
The numerator, 5x+125x+12, is zero when 5x=125x = -12, which means x=125x = -\frac{12}{5}.
The denominator, x+1x+1, is zero when x=1x = -1.
So, our **critical points** are x=125x = -\frac{12}{5} and x=1x = -1.

STEP 6

Our **critical points** divide the number line into three intervals: (,125)(-\infty, -\frac{12}{5}), (125,1)(-\frac{12}{5}, -1), and (1,)(-1, \infty).
Let's test a point in each interval to see if the inequality holds true.

STEP 7

* **Interval 1:** (,125)(-\infty, -\frac{12}{5}).
Let's pick x=3x = -3.
Plugging this into 5x+12x+1\frac{5x+12}{x+1}, we get 5(3)+123+1=32=32\frac{5(-3)+12}{-3+1} = \frac{-3}{-2} = \frac{3}{2}, which is *not* less than zero.
So, this interval *doesn't* work.

STEP 8

* **Interval 2:** (125,1)(-\frac{12}{5}, -1).
Let's pick x=2x = -2.
Plugging this in, we get 5(2)+122+1=21=2\frac{5(-2)+12}{-2+1} = \frac{2}{-1} = -2, which *is* less than zero.
This interval works!

STEP 9

* **Interval 3:** (1,)(-1, \infty).
Let's pick x=0x = 0.
Plugging this in, we get 5(0)+120+1=121=12\frac{5(0)+12}{0+1} = \frac{12}{1} = 12, which is *not* less than zero.
So, this interval *doesn't* work.

STEP 10

The solution to the inequality is the interval (125,1)(-\frac{12}{5}, -1).

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