Domain of a Function: Multiple domains

The domain of the equation is the set of domain values (of the variable in the equation) for which all algebraic expressions in the equation are well defined,

From this, of course - we exclude numbers for which arithmetic operations are not defined,

In the expression on the left side of the given equation:

$9(4x-\frac{5}{x})=20(3x-\frac{6}{x+1})$

There is a multiplication operation between fractions whose denominators contain algebraic expressions that include the variable of the equation,

These fractions are considered defined as long as the expressions in their denominators are not equal to zero (since division by zero is not possible),

Therefore, the domain of definition of the variable in the equation will be obtained from the requirement that these expressions (in the denominators of the fractions) do not equal zero, as shown below:

For the fraction inside of the parentheses in the expression on the left side we obtain the following:

$\boxed{ x\neq0}$

For the fraction inside of the parentheses in the expression on the right side we obtain the following:

$x+1\neq0 \\$Proceed to solve the second inequality above (in the same way as solving an equation):

$x+1\neq0 \\ \boxed{x\neq-1}$

Therefore, the correct answer is answer A.

Note:

It should be noted that the above inequality is a point inequality and not a trend inequality (meaning it negates equality: $(\neq)$and does not require a trend: (<,>,\leq,\geq) ) which is solved exactly like solving an equation. This is unlike solving a trend inequality where different solution rules apply depending on the type of expressions in the inequality, for example: solving a first-degree inequality with one variable (which has only first-degree and lower algebraic expressions), is solved almost identically to solving an equation. However any division or multiplication operation of both sides by a negative number requires that the trend be revered.

To find the domain of the given function, we need to determine where the function is undefined due to division by zero. The function in question is:

$\frac{14}{x} - 6x = \frac{2}{x-5}$

We identify two fractions: $\frac{14}{x}$ and $\frac{2}{x-5}$. Each fraction has a denominator that can potentially cause division by zero:

• For $\frac{14}{x}$, the denominator $x$ shouldn't be zero. Thus, $x \neq 0$.
• For $\frac{2}{x-5}$, the denominator $x-5$ shouldn't be zero. Thus, $x \neq 5$.

By excluding these values from the set of all real numbers, we obtain the domain of the function. Therefore, the domain consists of all real numbers except for $x = 0$ and $x = 5$.

Thus, the domain of the function is $x \neq 0, x \neq 5$.

The given function is $\frac{10x+2}{\sqrt{x^2-4}}$. Finding its domain requires ensuring the denominator is not zero and the expression under the square root is positive.

First, identify the points where the expression under the square root is zero: set $x^2 - 4 = 0$.

Solving $x^2 - 4 = 0$:

• Add 4 to both sides: $x^2 = 4$
• Take the square root of both sides: $x = \pm 2$

This means the points $x = 2$ and $x = -2$ need further inspection since they make the expression zero (hence the denominator would be undefined).

Next, determine where $x^2 - 4 > 0$. This inequality can be rewritten as:

$(x - 2)(x + 2) > 0$

Evaluate the intervals determined by these critical points:

• Interval $x < -2$: Choose $x = -3$, then $(-3-2)(-3+2) = (-5)(-1) = 5 > 0$
• Interval $-2 < x < 2$: Choose $x = 0$, then $(0-2)(0+2) = (-2)(2) = -4 < 0$
• Interval $x > 2$: Choose $x = 3$, then $(3-2)(3+2) = (1)(5) = 5 > 0$

Therefore, the expression is positive in the intervals $x < -2$ and $x > 2$.

To avoid the denominator being zero, these points are not included in the domain, confirming the domain as $x > 2$ and $x < -2$.

Thus, the solution to the problem is $x > 2, x < -2$, which corresponds to choice .

To find the domain of the function $\frac{8x}{\sqrt{2x^2-2}}$, we need to ensure the denominator is not zero and is defined.

Since the denominator is $\sqrt{2x^2 - 2}$, we have the condition:

• $2x^2 - 2 > 0$

Let's solve the inequality $2x^2 - 2 > 0$.

First, set the equation to zero to find critical points:

$2x^2 - 2 = 0$

Simplify and solve for $x$:

$2x^2 = 2$

$x^2 = 1$

$x = \pm 1$

The critical points divide the number line into three intervals: $x < -1$, $-1 < x < 1$, and $x > 1$.

We need to test these intervals to see where $2x^2 - 2 > 0$.

• For $x < -1$, choose $x = -2$:
• $2(-2)^2 - 2 = 8 - 2 = 6$ (positive)

• For $-1 < x < 1$, choose $x = 0$:
• $2(0)^2 - 2 = -2$ (negative)

• For $x > 1$, choose $x = 2$:
• $2(2)^2 - 2 = 8 - 2 = 6$ (positive)

Therefore, the intervals where $2x^2 - 2 > 0$ are $x < -1$ or $x > 1$.

Thus, the domain of the function is $x > 1$ or $x < -1$, in interval notation this is $(-\infty, -1) \cup (1, \infty)$.

So, the correct choice is $x > 1, x < -1$, corresponding to choice 4.

Therefore, the domain of the function is $x > 1, x < -1$.

To solve this problem, we must determine the domain of the function $\frac{x+2}{\sqrt{3x^2-9}}$.

To begin, the expression inside the square root, $3x^2 - 9$, must be greater than 0 for the square root to be real and the function to be defined. Thus, we set up the inequality:

$3x^2 - 9 > 0$

Next, solve this inequality for $x$:

• Start by factoring: $3x^2 - 9 = 3(x^2 - 3)$
• Set the factor equal to zero to find the critical points: $x^2 - 3 = 0$
• Solve for $x$:
• $x^2 = 3$
• $x = \pm \sqrt{3}$

Now, determine the intervals where $3x^2 - 9$ is positive. Consider the intervals defined by the critical points $x = \sqrt{3}$ and $x = -\sqrt{3}$:

• Interval 1: $(-\infty, -\sqrt{3})$
• Interval 2: $(- \sqrt{3}, \sqrt{3})$
• Interval 3: $(\sqrt{3}, \infty)$

Test a value from each interval in the inequality $3x^2 - 9 > 0$:

• For interval 1, test $x = -2$ :
• $3(-2)^2 - 9 = 12 - 9 = 3 > 0$ (True)
• For interval 2, test $x = 0$ :
• $3(0)^2 - 9 = -9 < 0$ (False)
• For interval 3, test $x = 2$ :
• $3(2)^2 - 9 = 12 - 9 = 3 > 0$ (True)

Thus, the function is defined for $x$ in the intervals $(-\infty, -\sqrt{3})$ and $(\sqrt{3}, \infty)$.

This means that the domain of the function is:

$x>\sqrt{3},x<-\sqrt{3}$

Therefore, the solution to the problem is $x>\sqrt{3},x<-\sqrt{3}$.

To solve this problem, we'll determine the domain of the function $f(x) = \frac{\sqrt{x^2+2}}{3}$.

First, consider the expression inside the square root, $x^2 + 2$. In order for the square root to be defined for real numbers, the expression $x^2 + 2$ must be non-negative.

Let's analyze $x^2 + 2$:

• For any real number $x$, the expression $x^2$ is always non-negative.
• Adding 2 to $x^2$ means $x^2 + 2$ is always greater than or equal to 2.
• Thus, $x^2 + 2 \geq 2 > 0$ for all real numbers $x$.

Since the value under the square root is always positive for all real numbers, the square root, and hence the function $f(x)$, is defined for all real numbers.

Therefore, the function has no restrictions on its domain other than the real number system itself. There are no variables in the denominator that can make it zero, as it is the constant 3.

Thus, the domain of the function is all real numbers.

The correct answer choice is: All real numbers.

To determine the domain of the function $\frac{\sqrt{4x^2 - 8}}{5}$, we must ensure the expression under the square root is non-negative. This condition will make the function well-defined over the real numbers.

Steps to solve for the domain:

• Step 1: Set the expression inside the square root greater or equal to zero: $4x^2 - 8 \geq 0$.
• Step 2: Solve the inequality.

To solve the inequality $4x^2 - 8 \geq 0$:

Step 3: Add 8 to both sides:

$4x^2 \geq 8$

Step 4: Divide each term by 4 to simplify:

$x^2 \geq 2$

Step 5: Solve for $x$. When an inequality involves a square, interpret it as involving two cases. Thus, $x \geq \sqrt{2}$ OR $x \leq -\sqrt{2}$.

This inequality describes the values of $x$ for which the function is defined. These constitute the domain of the function. Therefore, the domain is $x \geq \sqrt{2}$ or $x \leq -\sqrt{2}$.

The correct answer choice is:

$x\ge\sqrt{2},x\le-\sqrt{2}$

To determine the domain of the function $\frac{\sqrt{4x^2-4}}{10}$, we need to ensure that the expression inside the square root is non-negative. This ensures the function is defined for those values of $x$.

First, set the expression inside the square root to be non-negative:

• $4x^2 - 4 \geq 0$

Next, solve this inequality:

• Factor the expression: $4(x^2 - 1) \geq 0$, which simplifies to $x^2 - 1 \geq 0$.
• Further factorization gives: $(x - 1)(x + 1) \geq 0$.

Now, determine the intervals where this product is non-negative:

• The critical points are $x = 1$ and $x = -1$.
• Test intervals determined by these critical points:
  • Interval $x < -1$: Choose $x = -2$, $(x - 1)(x + 1) = (-3)(-1) = 3 \geq 0$.
  • Interval $-1 \le x \le 1$: Choose $x = 0$, $(x - 1)(x + 1) = (-1)(1) = -1$ which is not $\geq 0$.
  • Interval $x > 1$: Choose $x = 2$, $(x - 1)(x + 1) = (1)(3) = 3 \geq 0$.

Therefore, the solution to the inequality $(x - 1)(x + 1) \geq 0$ is $x \le -1$ or $x \ge 1$.

Thus, the domain of the function is $x \ge 1$ or $x \le -1$.

In the context of the given choices, the solution corresponds to choice 4: $x \ge 1, x \le -1$.

The domain of the function is $x \ge 1$ or $x \le -1$.

To determine the domain of the function $\frac{\sqrt{3x^2+3}}{9}$, we need to ensure that the expression under the square root is non-negative:

$3x^2 + 3 \geq 0$

Simplifying this inequality, we can factor it:

• Factor out the common term: $3(x^2 + 1) \geq 0$.
• Since $3$ is a positive constant, we focus on $x^2 + 1 \geq 0$.
• The term $x^2$ is always non-negative, hence $x^2 + 1$ is always positive for any real number $x$, as the smallest value it can take, when $x = 0$, is 1.

Thus, $x^2 + 1$ is never negative, making the expression under the square root always non-negative.

Therefore, the domain of the function is all real numbers.

Emphasizing the conclusion: The entire domain of this function is all real numbers.

To solve this problem, first, we determine the condition under the square root function by solving:

$2.5x^2 - 5 \geq 0$.

This inequality ensures that the expression inside the square root is non-negative, a requirement for the square root function to be defined over real numbers.

• Step 1: Simplify the inequality:
• First, add 5 to both sides to isolate the term involving $x$: $2.5x^2 \geq 5$
• Step 2: Solve for $x^2$ by dividing both sides by 2.5: $x^2 \geq \frac{5}{2.5}$
• Step 3: Simplify the fraction: $x^2 \geq 2$
• Step 4: Solve for $x$ by taking the square root of both sides, considering positive and negative solutions: $x \geq \sqrt{2} \quad \text{or} \quad x \leq -\sqrt{2}$

These conditions define the interval for which the original function is defined, corresponding to the original prompt requirement of a non-negative under-the-root value.

Thus, the domain of the function $\frac{\sqrt{2.5x^2-5}}{5}$ is $x \ge \sqrt{2}$ or $x \le -\sqrt{2}$.

The correct choice among the provided options is:

$x \ge \sqrt{2}, x \le -\sqrt{2}$

To find the domain of the function $\frac{\sqrt{3x^2+7}}{12}$, we must ensure that the function is defined for all real numbers.

Step 1: Evaluate the expression under the square root, $3x^2 + 7$, which must be non-negative. Since it's a quadratic expression in the form of $ax^2 + b$, compute for any potential zero or negative range.

Step 2: Notice that $3x^2 + 7 \geq 0$ for all $x$ because $3x^2 \geq 0$ for any real number $x$ and adding 7 makes this entire expression always positive (i.e., tends upwards away from zero).

Step 3: As the denominator $12$ is a positive constant, it imposes no additional restrictions on the domain. Thus, the function is defined wherever the numerator is defined.

Conclusion: Since there's no $x$ that makes $3x^2 + 7 < 0$, the function is defined for all real numbers.

This means the domain of the function is all real numbers, confirmed by choice number 1: $\text{All real numbers}$.

To find the domain of the function $\frac{\sqrt{5x^2+2}}{10}$, we need to ensure that the expression under the square root is non-negative.

Let's examine the inequality:

This is always true since the expression $5x^2$ (a non-negative value for all real $x$) added to 2 will always be greater than or equal to zero. Consequently, $5x^2 + 2$ never takes a negative value, confirming the square root is always defined.

Therefore, the function $\frac{\sqrt{5x^2+2}}{10}$ is defined for all real numbers.

In conclusion, the domain of the function is all real numbers.

To find the domain of the rational equation $\frac{5x}{2(x-7)} = \frac{10}{8x}$, we need to ensure neither denominator equals zero.

Start by examining the first denominator, $2(x-7)$:

• Set $2(x-7) = 0$.
• Solve for $x$ to find $x - 7 = 0$, resulting in $x = 7$.

Next, examine the second denominator, $8x$:

• Set $8x = 0$.
• Solve for $x$ to find $x = 0$.

Therefore, the function is undefined at $x = 0$ and $x = 7$. These values should be excluded from the domain.

Thus, the domain of the given rational equation is all real numbers except where $x = 0$ and $x = 7$.

This corresponds to the correct answer choice: $x \neq 0, x \neq 7$.

To solve this problem, we'll determine where the given expression is undefined:

• Step 1: Identify where the first fraction, $\frac{4}{x-2}$, is undefined. This fraction is undefined when its denominator is zero: $x-2 = 0$. Thus, $x = 2$.
• Step 2: Identify where the second fraction, $\frac{7x}{x-6}$, is undefined. This fraction is undefined when its denominator is zero: $x-6 = 0$. Thus, $x = 6$.
• Step 3: The expression is undefined at $x = 2$ and $x = 6$.

Therefore, the domain of the expression excludes $x = 2$ and $x = 6$.

The correct domain restriction is $x \neq 2, x \neq 6$.

The domain of the equation is the set of domain values (of the variable in the equation) for which all algebraic expressions in the equation are well defined,

From this, of course - we exclude numbers for which arithmetic operations are not defined,

In the expression on the left side of the given equation:

$\frac{7}{x+5}=\frac{6}{13x}$

There is a multiplication operation between fractions whose denominators contain algebraic expressions that include the variable of the equation.

These fractions are considered to be defined as long as the expression in their denominators is not equal to zero (given that division by zero is not possible),

Therefore, the domain of definition of the variable in the equation will be obtained from the requirement that these expressions (in the denominators of the fractions) do not equal zero, as follows:

For the fraction in the expression on the left side we obtain:

$x+5\neq0 \\$For the fraction in the expression on the right side we obtain:

$13x\neq0$

$$We will solve these inequalities (in the same way as solving an equation):

$x+5\neq0 \\ \boxed{x\neq-5}$

$13x\neq0 \hspace{8pt}\text{/:13} \\ \boxed{x\neq0}$$$

Therefore, the correct answer is answer A.

Note:

It should be noted that the above inequality is a point inequality and not a directional inequality (meaning it negates equality: $(\neq)$and does not require direction: (<,>,\leq,\geq) ) which is solved exactly like solving an equation. This is unlike solving a directional inequality where different solution rules apply depending on the type of expressions in the inequality. For example: solving a first-degree inequality with one variable (which only has first-degree algebraic expressions and below), is solved almost identically to solving an equation. However, any division or multiplication of both sides by a negative number requires reversing the direction.

To solve the problem, follow these steps:

• Step 1: Identify where each denominator is zero to find the domain restrictions.

• Step 2: Solve each condition separately to exclude the non-permissible $x$ values.

Now, let's work through each step:

Step 1: The first expression involves the denominator $5x - 6$. Set it to zero:

$5x - 6 = 0$

Solve for $x$:
$5x = 6$
$x = \frac{6}{5} = 1\frac{1}{5}$

This means the function is undefined for $x = 1\frac{1}{5}$.

Step 2: The second expression involves the denominator $x - 1$. Set it to zero:

$x - 1 = 0$

Solve for $x$:
$x = 1$

This means the function is undefined for $x = 1$.

The domain of this expression is all real numbers except where these denominators are zero. Therefore, the domain restriction is:

The values of $x$ cannot equal 1 or $1\frac{1}{5}$, which corresponds to choice 3.

Therefore, the solution to the problem is $x \neq 1, x \neq 1\frac{1}{5}$.

To solve this problem, we need to identify the values of $y$ for which the denominator of the expression becomes zero, as these values are not part of the domain.

First, let's simplify the denominator of the given equation:

Original equation: $\frac{\sqrt{15} + \frac{34}{z}}{4y - 12 + \frac{8}{2}} = 5$

Simplifying the terms: $34:z \text{ remains as it is for simplification purposes, and } \frac{8}{2} = 4$

Thus, the denominator becomes: $(4y - 12 + 4) = 4y - 8$

We need to ensure the denominator is not zero to avoid undefined expressions: $4y - 8 \neq 0$

Simplify and solve for $y$: $4y - 8 \neq 0 \implies 4y \neq 8 \implies y \neq 2$

Therefore, the equation is undefined for $y = 2$, and the answer is that the field of application excludes $y = 2$.

Given the possible choices for the problem, the correct choice is: $y\operatorname{\ne}2$

The solution to this problem is $y \neq 2$.

To solve this problem, we'll follow these steps:

• Step 1: Recognize and factor the expression in the numerator.
• Step 2: Simplify the fraction by canceling common factors.
• Step 3: Determine restrictions on the variable $x$.
• Step 4: Analyze if and when the equation holds true.

Now, let's work through each step:

Step 1: The numerator $x^2 - 81$ can be factored as a difference of squares: $(x-9)(x+9)$.

Step 2: Substitute this factorization into the equation:
$\frac{(x-9)(x+9)}{(x-9)(x+9)} = 1$.

Step 3: Simplify the fraction by canceling the common terms, giving $1 = 1$, which is always true, except where the expression is undefined.

Step 4: The expression is undefined when the denominator is zero, i.e., when $x - 9 = 0$ or $x + 9 = 0$. Thus, $x \neq 9$ and $x \neq -9$.

In conclusion, the given equation $\frac{x^2-81}{(x-9)(x+9)}=1$ is True only when $x \ne \pm9$.