II. Graph and construct a sign chart for the following rational inequalities.

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Ask by Harmon Ramos. in the Philippines

Dec 08,2024

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II. Graph and construct a sign chart for the following rational inequalities. $$ \frac{x^{2}+4 x+3}{x-1}>0 $$ Solution:

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To solve the rational inequality $$ \frac{x^{2}+4x+3}{x-1} > 0 $$, we will follow these steps: 1. **Factor the numerator**. 2. **Identify the critical points**. 3. **Determine the sign of the expression in each interval**. 4. **Construct a sign chart**. ### Step 1: Factor the numerator The numerator $$ x^2 + 4x + 3 $$ can be factored as follows: $$ x^2 + 4x + 3 = (x + 1)(x + 3) $$ So, we can rewrite the inequality as: $$ \frac{(x + 1)(x + 3)}{x - 1} > 0 $$ ### Step 2: Identify the critical points The critical points occur where the numerator is zero or where the denominator is zero. - **Numerator**: Set $$ (x + 1)(x + 3) = 0 $$: - $$ x + 1 = 0 $$ → $$ x = -1 $$ - $$ x + 3 = 0 $$ → $$ x = -3 $$ - **Denominator**: Set $$ x - 1 = 0 $$: - $$ x = 1 $$ Thus, the critical points are $$ x = -3, -1, 1 $$. ### Step 3: Determine the sign of the expression in each interval The critical points divide the number line into the following intervals: 1. $$ (-\infty, -3) $$ 2. $$ (-3, -1) $$ 3. $$ (-1, 1) $$ 4. $$ (1, \infty) $$ We will test a point from each interval to determine the sign of the expression $$ \frac{(x + 1)(x + 3)}{x - 1} $$. - **Interval $$ (-\infty, -3) $$**: Choose $$ x = -4 $$ $$ \frac{(-4 + 1)(-4 + 3)}{-4 - 1} = \frac{(-3)(-1)}{-5} = \frac{3}{-5} < 0 $$ - **Interval $$ (-3, -1) $$**: Choose $$ x = -2 $$ $$ \frac{(-2 + 1)(-2 + 3)}{-2 - 1} = \frac{(-1)(1)}{-3} = \frac{-1}{-3} > 0 $$ - **Interval $$ (-1, 1) $$**: Choose $$ x = 0 $$ $$ \frac{(0 + 1)(0 + 3)}{0 - 1} = \frac{(1)(3)}{-1} = \frac{3}{-1} < 0 $$ - **Interval $$ (1, \infty) $$**: Choose $$ x = 2 $$ $$ \frac{(2 + 1)(2 + 3)}{2 - 1} = \frac{(3)(5)}{1} = 15 > 0 $$ ### Step 4: Construct a sign chart Now we can summarize the signs in a sign chart: $$ \begin{array}{c|c|c|c|c} \text{Interval} & (-\infty, -3) & (-3, -1) & (-1, 1) & (1, \infty) \ \hline \text{Sign} & - & + & - & + \ \end{array} $$ ### Step 5: Determine the solution set The inequality $$ \frac{(x + 1)(x + 3)}{x - 1} > 0 $$ is satisfied in the intervals where the expression is positive: - From the sign chart, the solution is: - $$ (-3, -1) $$ - $$ (1, \infty) $$ ### Final Answer Thus, the solution to the inequality $$ \frac{x^{2}+4x+3}{x-1}>0 $$ is: $$ \boxed{(-3, -1) \cup (1, \infty)} $$ The solution to the inequality $$ \frac{x^{2}+4x+3}{x-1}>0 $$ is $$ (-3, -1) \cup (1, \infty) $$.

II. Graph and construct a sign chart for the following rational inequalities.

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II. Graph and construct a sign chart for the following rational inequalities. Solution: