Problem 63 - College Algebra

Question

In Exercises \(61-64,\) find the domain of each function. $$ f(x)-\sqrt{\frac{2 x}{x+1}-1} $$

Step-by-Step Solution

Answer

The domain of the function is \([1, +\infty)\)

Step 1: Find the Condition for the Square Root

We need to find values of \(x\) for which the argument of the square root is non-negative. This leads us to the inequality \(\frac{2 x}{x+1}-1 \geq 0\).

Step 2: Solve the Inequality

To solve the inequality, we first get rid of the fraction by multiplying all terms by \(x+1\), then we simplify and solve for \(x\). This gives us \(2x-(x+1) \geq 0\), simplifying further we get \(x-1 \geq 0\), which implies \(x \geq 1\)

Step 3: Find the Condition for the Rational Expression

We need to find for which values of \(x\), the denominator of the rational expression \(x+1\) is not equal to zero. The denominator \(x+1\) equals zero if \(x = -1\). Therefore, \(x\) should not be equal to -1 for the rational expression inside the square root to be defined.

Step 4: Combine the Conditions for the Domain

Combining the two conditions, we conclude that all values \(x \geq 1\) and \(x\) not equal to -1 satisfy both conditions. Since -1 is not included in the interval \([1, +\infty)\), we do not need to exclude it separately. Therefore, the domain of the function is \([1, +\infty)\)

Key Concepts

Square Root Inequality

In mathematics, handling square root expressions is crucial to defining valid values. Specifically, when dealing with a square root, the expression inside it—that is, the radicand—must be non-negative to produce a real number. This is because the square root of a negative number is not defined in real numbers.

Let's consider the problem where you have an inequality like \( \sqrt{\frac{2x}{x+1} - 1} \). To find its domain, you need to ensure that:

• The expression inside the square root, \( \frac{2x}{x+1} - 1 \), is greater than or equal to zero.
• This leads to solving \( \frac{2x}{x+1} - 1 \geq 0 \).

Once you solve this inequality, you will find that only certain values of \( x \) make it non-negative, ensuring the square root is defined.

Rational Expression

Rational expressions are fractions where both the numerator and the denominator are polynomials. They are significant because they often emerge in functions that model real-world phenomena.

Analyzing the rational part of the function, \( \frac{2x}{x+1} \), requires caution. The denominator, \( x+1 \), must never be zero, as division by zero is undefined. Identify values of \( x \) where the denominator becomes zero, and exclude these from the function's domain.

In this exercise, \( x+1 = 0 \) when \( x = -1 \). Therefore, \( x=-1 \) must be excluded from the domain of the function. For any rational expression, remember:

• The denominator should never be zero.
• Exclude any values causing the denominator to be zero from the domain.

Inequality Solving

Solving inequalities is fundamental in determining allowable values of \( x \) within an expression. Begin by treating the inequality similar to an equation where operations are performed to isolate the variable of interest.

For \( \frac{2x}{x+1} - 1 \geq 0 \), you'd first get rid of the fraction. Multiply both sides by \( x+1 \) (assuming \( x+1 eq 0 \)):

• The inequality becomes \( 2x - (x+1) \geq 0 \).
• Simplify it to \( x - 1 \geq 0 \), thus \( x \geq 1 \).

This solution indicates that \( x \) must be greater than or equal to 1 to keep the expression valid. Always ensure each manipulation of an inequality considers both equality and strict inequality where appropriate.

Interval Notation

Interval notation is a convenient way to express a range of values that an unknown variable like \( x \) can take. It's concise and clear, displaying a range through parentheses or brackets.

Using interval notation, one writes \( [1, \infty) \) to indicate all values from 1 to infinity, including 1 but not including infinity.

• Brackets \([ \ ]\) signify that the endpoint is included.
• Parentheses \((\ )\) mean the endpoint is not included.

For the problem you tackled, solving \( x \geq 1 \) and ensuring \( x eq -1 \) was inherent, hence -1 is outside the domain naturally leaving us with the interval \([1, \infty)\). When writing intervals, always use brackets and parentheses appropriate to the solution.