Domain of a Function: Finding the Domain of a Square Root Function

To determine the domain of the function $\frac{2x+20}{\sqrt{2x-10}}$, we must ensure that the expression under the square root is non-negative, because the square root of a negative number is not defined in the real numbers.

We start by analyzing the denominator, specifically the square root, $\sqrt{2x-10}$. For the square root to be valid (for real numbers), we require:

• $2x-10 \geq 0$

Now, solve the inequality $2x - 10 \geq 0$:

• Add 10 to both sides: $2x \geq 10$
• Divide both sides by 2: $x \geq 5$

However, since the expression $2x-10$ also prohibits zero in the denominator (as the square root in the denominator cannot be zero), we strictly have:

• $x > 5$

Thus, the domain of the function is all $x$ such that $x > 5$.

Therefore, the domain of the function $\frac{2x+20}{\sqrt{2x-10}}$ is $x > 5$.

To determine the domain of the function $\frac{23}{\sqrt{x}}$, we must ensure the function is defined for all values in its domain. The expression involves a square root and a division.

• First, consider the square root, $\sqrt{x}$. This is only defined for $x \geq 0$. Therefore, initially, $x$ must be non-negative.

• Second, because the square root is in the denominator of a fraction, $\sqrt{x}$ must not equal zero to avoid division by zero. Thus, $x$ must be strictly greater than 0.

Combining these conditions, we find that the domain of the function is $x > 0$.

Therefore, the domain of the function is $x > 0$, which corresponds to choice 3 from the provided options.

To determine the domain of the function $\frac{3}{\sqrt{x-2}}$, we need to consider the constraints imposed by the square root and the fraction.

First, the expression inside the square root must be non-negative: $x - 2 \geq 0$. This simplifies to:

• $x \geq 2$

However, because the expression $\sqrt{x-2}$ is in the denominator of the fraction, we must also ensure that it is not equal to zero, as division by zero is undefined. Therefore, we have:

• $x - 2 \neq 0$

Solving this inequality gives:

• $x \neq 2$

Together, the solution to both conditions is that $x$ must be greater than 2 to ensure the function is defined. Thus, the domain of the function is:

$x > 2$

Therefore, the correct choice for the domain is $x > 2$.

To solve this problem, we need to determine the domain of the function given by $\frac{3}{\sqrt{2x-3}}$. The fraction is undefined whenever the denominator is zero, and the square root requires the expression inside to be positive.

Let's break down the steps:

• Since the square root is in the denominator, $2x - 3$ must be strictly greater than zero.
• Set up the inequality: $2x - 3 > 0$.
• Solve the inequality for $x$:
• Add 3 to both sides: $2x > 3$.
• Divide both sides by 2: $x > 1.5$.

The domain of the function is all real numbers $x$ such that $x > 1.5$. Therefore, we write the domain as $x > 1.5$.

Hence, the correct choice among the given options is: $x > 1.5$ which corresponds to choice number 3.

To solve the problem of finding the domain of the function $\frac{20}{\sqrt{4x-2}}$, we need to ensure that the expression under the square root is non-negative, and that we do not divide by zero.

**Step 1:** Solve for when the expression inside the square root is non-negative:

$4x - 2 \geq 0$

Add 2 to both sides:

$4x \geq 2$

Divide both sides by 4:

$x \geq 0.5$

**Step 2:** Ensure the denominator is not zero:

$4x - 2 \neq 0$

From $4x - 2 = 0$, we solve:

$4x = 2$

$x = 0.5$

Since at $x = 0.5$, the denominator becomes zero, we exclude this point. Therefore, x > 0.5 .

To solve this problem, we need to determine the domain of the function given by:

$f(x) = \frac{4}{\sqrt{x+8}}$

We must ensure that the function is defined for all $x$. This involves considering the conditions under which the square root is valid and the denominator is non-zero.

Step 1: Analyze the square root expression $\sqrt{x+8}$. The expression inside the square root must be non-negative:

$x + 8 \ge 0$

Step 2: Solve the inequality:

• Subtract 8 from both sides: $x \ge -8$

Step 3: Consider the division by zero issue. The denominator $\sqrt{x+8}$ must be strictly greater than zero to avoid division by zero. Thus, we adjust the inequality to:

$x + 8 > 0$

Step 4: Solve the second inequality:

• Subtract 8 from both sides: $x > -8$

Thus, the domain of the function is all $x$ such that $x > -8$.

Review of the answer choices shows that the correct choice, consistent with our findings, is:

$x > -8$

Therefore, the domain of the function is $x > -8$.

To find the domain of the function $\frac{12}{\sqrt{4x-4}}$, let's analyze the conditions necessary for the function to be defined.

The expression under the square root, $4x - 4$, must be positive, as the square root of a negative number is not defined in the real numbers, and division by zero is undefined. Therefore, we need:

• $4x - 4 > 0$

Solve this inequality step by step:

• Add 4 to both sides: $4x - 4 + 4 > 0 + 4$, which simplifies to $4x > 4$.
• Divide both sides by 4: $\frac{4x}{4} > \frac{4}{4}$, which simplifies to $x > 1$.

The inequality $x > 1$ describes the domain of the function.

Therefore, the domain of the function $\frac{12}{\sqrt{4x-4}}$ is $x > 1$.

To find the domain of the function $\frac{2x+2}{\sqrt{x+2.5}}$, we need to determine for which values of $x$ the expression is defined.

Step 1: Identify the restriction on the square root.
The square root function $\sqrt{x+2.5}$ is defined when the expression inside the square root is non-negative. Thus, we have the inequality:

$x + 2.5 \geq 0$

Step 2: Solve the inequality for $x$.
Subtract 2.5 from both sides:

$x \geq -2.5$

Step 3: Ensure the denominator is not zero because division by zero is undefined.
Since $x + 2.5 \neq 0$, we require:

$x \neq -2.5$

Therefore, combining these results, the domain of the function is:

$x > -2.5$

The correct answer to the problem, represented as a choice, is:

x > -2.5

To find the domain of the function $f(x) = \frac{3}{\sqrt{x-10}}$, follow these steps:

• First, ensure the expression under the square root, $x-10$, is non-negative. This gives $x-10 \geq 0$, or equivalently, $x \geq 10$.
• Second, the denominator of the function, $\sqrt{x-10}$, cannot be zero. This means $\sqrt{x-10} \neq 0$, leading to $x-10 \neq 0$. Therefore, $x \neq 10$.

Combining these conditions, the value of $x$ must satisfy $x > 10$. This ensures both the definition of the square root and the non-zero nature of the denominator.

The correct domain of the function is $x > 10$.

To determine the domain of the function $\frac{5}{\sqrt{x-5}}$, we must ensure that the expression inside the square root, $x-5$, is positive. Furthermore, because the square root is in the denominator, $x-5$ must be greater than zero:

• Step 1: Set the argument of the square root greater than zero: $x-5 > 0$.
• Step 2: Solve the inequality: Add 5 to both sides to get $x > 5$.

Since the inequality $x > 5$ ensures that the denominator is neither zero nor negative, it defines the domain of the function. Thus, the function $\frac{5}{\sqrt{x-5}}$ is defined for all real numbers $x$ where $x > 5$.

Therefore, the domain of the function is $x > 5$.

The function given is $\frac{3x+12}{\sqrt{5x-10}}$.

To find the domain, we focus on the expression within the denominator's square root: $\sqrt{5x-10}$.

The expression $5x-10$ must be greater than $0$ for the square root to be defined and the denominator to be non-zero.

Let's solve the inequality:

• Set $5x-10 > 0$.
• Add $10$ to both sides: $5x > 10$.
• Divide both sides by $5$: $x > 2$.

This means the domain of the function is all $x$ such that $x > 2$.

The domain is, therefore, correctly expressed as $x > 2$.

To find the domain of the function $\frac{4x-10}{\sqrt{2.5x-10}}$, we need to ensure the expression under the square root is positive since it cannot equal zero or be negative.

Step 1: Set up the inequality based on the square root:

$2.5x - 10 > 0$

Step 2: Solve the inequality for $x$:

• Add 10 to both sides: $2.5x > 10$
• Divide both sides by 2.5: $x > \frac{10}{2.5}$
• Calculate: $x > 4$

Step 3: Interpret the result:

The domain of the function is all real numbers greater than 4, $x > 4$, ensuring the expression inside the square root is always positive.

Thus, the correct domain is represented by choice 2: $x > 4$.

To determine the domain of the function $\frac{2x+2}{\sqrt{x-16}}$, follow these steps:

• Step 1: Identify the constraint imposed by the square root in the denominator.
• Step 2: Solve the inequality $x-16 > 0$.
• Step 3: Interpret the solution in terms of the domain.

Let's proceed:

Step 1: The function $\frac{2x+2}{\sqrt{x-16}}$ has a square root in the denominator. For the square root to be defined in the real number system and prevent division by zero, the expression under the square root, $x-16$, must be greater than zero.

Step 2: Solve the inequality:

$x - 16 > 0$

Add 16 to both sides:

$x > 16$

Step 3: The solution $x > 16$ means that the domain of the function is all real numbers greater than 16.

Therefore, the domain of the function is $x > 16$, which corresponds to choice 2.

To solve this problem, we need to determine where the function $f(x) = \frac{2x+2}{\sqrt{2x-8}}$ is defined.

For the fraction to be defined, the denominator cannot be zero, and for the square root to be defined, the radicand (the expression inside the square root) must be non-negative.

Therefore, we need to solve the inequality:

$2x - 8 \geq 0$

Solving this inequality involves the following steps:

• Add 8 to both sides: $2x \geq 8$
• Divide both sides by 2: $x \geq 4$

However, if $x = 4$, the expression inside the square root is zero, making the denominator zero and the overall expression undefined.

As a result, the domain of the function is $x > 4$.

Therefore, the domain of the function is $x > 4$.

To determine the domain of the function $\frac{x+7}{\sqrt{x-7}}$, we need to ensure the denominator remains defined and non-zero. As follows:

First, focus on the denominator $\sqrt{x-7}$. The expression under the square root, $x-7$, must be greater than zero for the square root to be defined and not produce zero in the denominator:

• $x - 7 > 0$

This simplifies to:

• $x > 7$

Since the expression under the square root must always be positive for this rational function to be defined, and $\sqrt{x-7}$ in the denominator implies it cannot equal zero, our analysis is complete. Consequently, the domain of the function is the set of all $x$ such that:

The domain of the function is $x > 7$.