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Chapter 2: Problem 63

Find the domain of each function. $$f(x)=\sqrt{\frac{2 x}{x+1}-1}$$

Short Answer

Expert verified
The domain of the function \(f(x) = \sqrt{\frac{2x}{x+1} -1}\) is \(x\) in (-鈭�, -1) U [1, 鈭�).

Step by step solution

01

Discard the undefined numbers for denominator

In our function, \(x+1\) is in the denominator. A function is undefined when the denominator is 0. So, \(x + 1 鈮� 0\). Solving for \(x\) gives \(x 鈮� -1\). This means \(x\) can be any real number except -1.
02

Set the expression under the square root 鈮� 0

The number under the square root must be greater than or equal to zero. Therefore, we set \(\frac{2x}{x+1} -1 鈮� 0\). Simplifying the inequality we obtain \(2x 鈮� x + 1\). Solving for \(x\) gives \(x 鈮� 1\). So, \(x\) can be any real number that is greater or equal to 1.
03

Find the intersection of the solutions from step 1 and 2

The domain of the function is the intersection of the solutions from step 1 and step 2. So \(x\) cannot be -1 but it must also be greater than or equal to 1, which makes the domain as \(x\) in (-鈭�, -1) U [1, 鈭�).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inequalities in Calculus
Understanding inequalities in calculus is crucial for finding the domain of functions, especially when dealing with square roots and rational expressions. Inequalities help determine where an expression is valid or defined. For example, if you have an inequality like \( \frac{2x}{x+1} - 1 \geq 0 \), you're looking at where the expression under the square root is non-negative.
This ensures that the expression remains within the set of real numbers, which is necessary for finding domains in real-valued functions.
  • First, rearrange and solve the inequality. This might involve clearing fractions or simplifying terms.
  • Next, find the value or range of \( x \) where the inequality holds true.
  • Finally, use this information alongside other constraints to find the required domain.
By combining these steps, you can effectively determine the intervals where your function is defined.
Square Root Function
The square root function has a unique property: it only produces real numbers when its argument is non-negative. This means that for a function like \( \sqrt{\frac{2x}{x+1} - 1} \), the expression \( \frac{2x}{x+1} - 1 \) must be greater than or equal to zero for the square root to exist.
This constraint ensures that the argument of the square root does not become negative, preventing undefined or imaginary results in the real number system. To handle such functions:
  • Identify the expression under the square root.
  • Set the expression greater than or equal to zero, since negative values won't provide real results.
  • Solve the inequality to find values of \( x \) that make the function valid.
These steps ensure that your function remains within the real number domain.
Rational Expressions
Rational expressions involve fractions that have polynomials in their denominators and numerators. They are defined everywhere except where their denominators are zero, as division by zero is undefined.
Consider an expression like \( \frac{2x}{x+1} \). Here, the denominator \( x+1 \) should not equal zero. This requires setting \( x+1 eq 0 \) and solving for \( x \), yielding \( x eq -1 \).
When dealing with rational expressions:
  • Always check where the denominator equals zero, as these points must be excluded.
  • Combine these exclusions with other conditions, such as those from square roots, to find the valid domain.
This approach helps you exclude invalid points and find the intersection of all conditions to define the domain accurately.

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Most popular questions from this chapter

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