Question

solve the inequality
\\(\frac{x + 4}{x + 1} - 3 < 0\\)
and write your answer in interval notation.
\\(\bigcirc (-\infty, -1) \cup (\frac{1}{2}, \infty)\\)
\\(\bigcirc (\frac{1}{2}, \infty)\\)
\\(\bigcirc (-\infty, -1)\\)
\\(\bigcirc (-\infty, -\frac{1}{2})\\)
\\(\bigcirc (-\infty, -3) \cup (3, \infty)\\)

Explanation:

Step1: Simplify the inequality

First, we combine the terms in the inequality \(\frac{x + 4}{x + 1}-3<0\). We get a common denominator of \(x + 1\):

$$ \frac{x + 4-3(x + 1)}{x + 1}<0 $$

Then, we expand and simplify the numerator:

$$ \frac{x + 4-3x - 3}{x + 1}<0\implies\frac{-2x + 1}{x + 1}<0 $$

Step2: Find critical points

The critical points occur where the numerator is zero or the denominator is zero.

• For the numerator \(-2x+1 = 0\), we solve for \(x\):

$$ -2x=-1\implies x=\frac{1}{2} $$

• For the denominator \(x + 1=0\), we solve for \(x\):

$$ x=-1 $$

Step3: Test intervals

We divide the number line into three intervals based on the critical points \(x=-1\) and \(x = \frac{1}{2}\):

1. Interval \((-\infty,-1)\): Let's choose a test point, say \(x=-2\). Substitute into \(\frac{-2x + 1}{x + 1}\):

$$ \frac{-2(-2)+1}{-2 + 1}=\frac{4 + 1}{-1}=-5<0 $$

So the inequality is satisfied in this interval.

1. Interval \((-1,\frac{1}{2})\): Let's choose a test point, say \(x = 0\). Substitute into \(\frac{-2x + 1}{x + 1}\):

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

So the inequality is not satisfied in this interval.

1. Interval \((\frac{1}{2},\infty)\): Let's choose a test point, say \(x = 1\). Substitute into \(\frac{-2x + 1}{x + 1}\):

$$ \frac{-2(1)+1}{1+1}=\frac{-1}{2}<0 $$

So the inequality is satisfied in this interval.

We also need to note that \(x=-1\) makes the denominator zero, so it is not included in the domain, and \(x=\frac{1}{2}\) makes the numerator zero, and at \(x = \frac{1}{2}\), the expression is zero, which does not satisfy the strict inequality \(<0\).

Answer:

\((-\infty,-1)\cup(\frac{1}{2},\infty)\)