Problem 47 - Precalculus

Question

Find the domain of the function. \(f(x)=\log \left(\frac{x+2}{x^{2}-1}\right)\)

Step-by-Step Solution

Answer

The domain of \(f(x)\) is \((-\infty, -2) \cup (-1, 1) \cup (1, \infty)\).

Step 1: Understand the Function Expression

The function is defined as a logarithm, \(f(x) = \log \left( \frac{x+2}{x^2-1} \right)\). For the logarithm to be defined, its argument must be positive, i.e., \(\frac{x+2}{x^2-1} > 0\). Additionally, the denominator cannot be zero.

Step 2: Solve the Inequality

We need to solve \(\frac{x+2}{x^2-1} > 0\). This occurs when \(x+2 > 0\) and \(x^2-1 > 0\) are of the same sign. Determine critical points where the expression changes sign.

Step 3: Find Critical Points

The critical points occur when either the numerator or the denominator is zero. \(x+2=0\) gives \(x=-2\). \(x^2-1=0\) is a difference of squares and factors to \((x-1)(x+1)=0\), giving \(x=1\) and \(x=-1\). These points divide the number line into intervals.

Step 4: Analyze Intervals

The intervals to consider are \((-\infty, -2)\), \((-2, -1)\), \((-1, 1)\), and \((1, \infty)\). Test points from each interval in the inequality to determine where the expression is positive.

Step 5: Include Endpoint Conditions

Check if the critical points \(x=-2\), \(x=-1\), and \(x=1\) are included or excluded based on the original conditions: \(x+2eq 0\), \(x^2-1 eq 0\). Thus, exclude \(x=-2\), \(x=1\), and \(x=-1\).

Step 6: Compile the Domain

From the above analysis, the intervals where \(\frac{x+2}{x^2-1}>0\) are \((-\infty, -2) \cup (-1, 1) \cup (1, \infty)\), none of which include the critical points. Combine this with the endpoint conditions to define the domain.

Key Concepts

Logarithmic Functions

Logarithmic functions serve as the inverse of exponential functions. In essence, they help us solve problems involving exponential growth or decay. The function given in the exercise, \(f(x) = \log \left( \frac{x+2}{x^2-1} \right)\), is a log function with the argument \(\frac{x+2}{x^2-1}\). For a logarithm to be valid, its argument must be greater than zero. This requirement ensures that the logarithm exists in the real number system because the log of zero or a negative number in the base-10 system is undefined. Hence, determining the domain of a logarithmic function involves finding when its argument is positive.

Inequalities

Inequalities are crucial in defining the domain of functions because they demonstrate where a function is valid. In this case, the function's argument \(\frac{x+2}{x^2-1}\) must satisfy the inequality \(\frac{x+2}{x^2-1} > 0\). This inequality is satisfied only when both the numerator \(x+2\) and the denominator \(x^2-1\) have the same sign.

• For \(x+2\) to be positive, \(x > -2\).
• For \(x^2-1\) to be positive, you need \(x > 1\) or \(x < -1\).

By exploring different intervals between the critical points, we can evaluate where this inequality holds true, leading us to the valid domain for the function.

Rational Expressions

Rational expressions involve a numerator and a denominator, both of which can be polynomial expressions. In the function \(\frac{x+2}{x^2-1}\), \(x+2\) is the numerator, and \(x^2-1\) is the denominator. To explore the domain of the function further, it's essential to check where the expression itself is undefined.
Rational expressions become undefined when their denominator equals zero. Therefore, setting \(x^2-1=0\) reveals the critical points where the expression could be undefined, such as \(x = -1\) and \(x = 1\). These points should be carefully considered as they may need to be excluded from the domain.

Critical Points

Critical points occur where the sign of an expression changes or where it becomes undefined. They are essential in determining the intervals over which a function is valid. For the function \(f(x) = \log \left( \frac{x+2}{x^2-1} \right)\), the critical points are found by setting \(x+2 = 0\) and \(x^2-1 = 0\).
Solving these equations gives critical points at \(x = -2\), \(x = -1\), and \(x = 1\). These points divide the number line and create intervals:

• \(( -\infty, -2 )\)
• \((-2, -1)\)
• \((-1, 1)\)
• \((1, \infty)\)

By plugging sample points from each interval into the inequality \(\frac{x+2}{x^2-1} > 0\), we can identify where the expression is positive and ascertain the valid domains.