Problem 58 - Precalculus: Functions and Graphs

Question

Solve the equation or inequality. Express the solutions in terms of intervals whenever possible. $$\frac{x+1}{x^{2}-25} \leq 0$$

Step-by-Step Solution

Answer

The solution is \((-\infty, -5) \cup [-1, 5)\).

Step 1: Identify Critical Points

To find the values where the inequality changes, set the numerator and the denominator to zero separately. The numerator \(x+1=0\) gives \(x=-1\). The denominator \(x^2 - 25 = 0\) gives \(x=5\) and \(x=-5\). Thus, the critical points are \(-5, -1,\) and \(5\).

Step 2: Determine Intervals

Using the critical points identified, split the real number line into intervals: \((-\infty, -5)\), \((-5, -1)\), \((-1, 5)\), and \((5, \infty)\).

Step 3: Evaluate Sign on Each Interval

Choose a test point from each interval to determine the sign of \(\frac{x+1}{x^2-25}\). - For \((-\infty, -5)\), use \(x=-6\): \(\frac{-6+1}{(-6)^2-25} = \frac{-5}{11}<0\). - For \((-5, -1)\), use \(x=-3\): \(\frac{-3+1}{(-3)^2-25}= \frac{-2}{-16}>0\).- For \((-1, 5)\), use \(x=0\): \(\frac{0+1}{0^2-25}= \frac{1}{-25}<0\).- For \((5, \infty)\), use \(x=6\): \(\frac{6+1}{6^2-25}= \frac{7}{11}>0\).

Step 4: Consider Boundary Points

Check the inequality at each critical point. At \(x=-5\) and \(x=5\), the denominator is zero, so the expression is undefined at these points. At \(x=-1\), the expression is zero which satisfies the inequality \(\leq 0\).

Step 5: Write Solution

The solution consists of intervals where the expression is less than or equal to zero: \((-\infty, -5)\cup [-1, 5)\). Include \(-1\) as it satisfies \(\leq 0\) and exclude \(-5\) and \(5\) as the expression is undefined there.

Key Concepts

Critical Points

In solving inequalities like \(\frac{x+1}{x^{2}-25} \leq 0\), finding critical points is essential. These points are where the expression might change direction, or become undefined. To identify them, you need to set both the numerator and the denominator equal to zero.

• Numerator: Solving \(x+1=0\) gives \(x=-1\).
• Denominator: Solve \(x^2-25=0\) or \((x+5)(x-5)=0\), yielding roots \(x=-5\) and \(x=5\). These points make the inequality undefined.

Thus, the critical points to consider are \(-5\), \(-1\), and \(5\). Identifying these points helps us later determine the intervals to test changes in sign.

Interval Notation

After identifying critical points, you use them to break down the real number line into intervals. This helps to determine the behavior of our inequality across different sections of the number line. For our inequality:- The critical points at \(-5\), \(-1\), and \(5\) divide the line into intervals: \((-\infty, -5), (-5, -1), (-1, 5),\) and \((5, \infty)\).Using interval notation is very efficient. It allows you to clearly express which intervals are included in the solution and what values are specifically excluded. It's important here because our final answer will indicate where the inequality is less than or equal to zero, highlighting where the values meet the inequality's conditions.

Numerator and Denominator

In an inequality fraction like \(\frac{x+1}{x^{2}-25}\), understanding the roles of the numerator and denominator is crucial.

• The numerator: This is \(x+1\). When it equals zero, the entire fraction is zero. This can be part of your solution if the inequality allows zero (\(\leq 0\)).
• The denominator: This is \(x^{2}-25\). It must not be zero because division by zero is undefined. So, solutions at \(x=-5\) and \(x=5\) are excluded from your interval.

Analyzing both separately helps to decipher which sections of the number line make the inequality true or false. A practical tip is to first find the points that make each zero, then test these points on the original inequality to understand their impact on the fraction's sign across different intervals.

Test Points

To determine the sign of the inequality within each interval, you choose test points. These are values chosen from each interval to understand the behavior of the inequality at those sections.For the inequality \(\frac{x+1}{x^{2}-25}\), you'd select points like so:

• In \((-\infty, -5)\), use \(x=-6\): The expression results in \(\frac{-5}{11} < 0\).
• In \((-5, -1)\), use \(x=-3\): The expression results in \(\frac{-2}{-16} > 0\).
• In \((-1, 5)\), use \(x=0\): The expression results in \(\frac{1}{-25} < 0\).
• In \((5, \infty)\), use \(x=6\): The expression results in \(\frac{7}{11} > 0\).

By observing the signs, you can accurately predict which intervals satisfy the original inequality. It’s a step-by-step checking process that ensures none of the possible intervals are overlooked, allowing you to verify your solution is correct.