Examples with solutions for Domain of a Function: Multiple domains

Exercise #1

Determine the area of the domain without solving the expression:

9(4x5x)=20(3x6x+1) 9(4x-\frac{5}{x})=20(3x-\frac{6}{x+1})

Video Solution

Step-by-Step Solution

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(4x5x)=20(3x6x+1) 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:

x0 \boxed{ x\neq0}

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

x+10 x+1\neq0 \\ Proceed to solve the second inequality above (in the same way as solving an equation):

x+10x1 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.

Answer

x0,x1 x≠0,x≠-1

Exercise #2

Determine the area of the domain without solving the expression:

7x+5=613x \frac{7}{x+5}=\frac{6}{13x}

Video Solution

Step-by-Step Solution

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:

7x+5=613x \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+50 x+5\neq0 \\ For the fraction in the expression on the right side we obtain:

13x0 13x\neq0

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

x+50x5 x+5\neq0 \\ \boxed{x\neq-5}

13x0/:13x0 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.

Answer

x0,x5 x≠0,x≠-5

Exercise #3

15+34:z4y12+8:2=5 \frac{\sqrt{15}+34:z}{4y-12+8:2}=5

What is the field of application of the equation?

Video Solution

Step-by-Step Solution

To solve this problem, we need to identify the values of yy 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: 15+34z4y12+82=5 \frac{\sqrt{15} + \frac{34}{z}}{4y - 12 + \frac{8}{2}} = 5

Simplifying the terms: 34:z remains as it is for simplification purposes, and 82=4 34:z \text{ remains as it is for simplification purposes, and } \frac{8}{2} = 4

Thus, the denominator becomes: (4y12+4)=4y8 (4y - 12 + 4) = 4y - 8

We need to ensure the denominator is not zero to avoid undefined expressions: 4y80 4y - 8 \neq 0

Simplify and solve for yy: 4y80    4y8    y2 4y - 8 \neq 0 \implies 4y \neq 8 \implies y \neq 2

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

Given the possible choices for the problem, the correct choice is: y2 y\operatorname{\ne}2

The solution to this problem is y2 y \neq 2 .

Answer

y2 y\operatorname{\ne}2

Exercise #4

Find the area of domain (no need to solve)

14x6x=2x5 \frac{14}{x}-6x=\frac{2}{x-5}

Video Solution

Answer

x0,x5 x≠0,x≠5

Exercise #5

Given the following function:

10x+2x24 \frac{10x+2}{\sqrt{x^2-4}}

What is the domain of the function?

Video Solution

Answer

x > 2,x < -2

Exercise #6

Look at the following function:

5x+4x29 \frac{5x+4}{\sqrt{x^2-9}}

What is the domain of the function?

Video Solution

Answer

x > 3,x < -3

Exercise #7

Look at the following function:

8x2x22 \frac{8x}{\sqrt{2x^2-2}}

What is the domain of the function?

Video Solution

Answer

x > 1,x < -1

Exercise #8

Look at the following function:

x+23x29 \frac{x+2}{\sqrt{3x^2-9}}

What is the domain of the function?

Video Solution

Answer

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

Exercise #9

Look at the following function:

x2+23 \frac{\sqrt{x^2+2}}{3}

What is the domain of the function?

Video Solution

Answer

All real numbers

Exercise #10

Look at the following function:

4x285 \frac{\sqrt{4x^2-8}}{5}

What is the domain of the function?

Video Solution

Answer

x2,x2 x\ge\sqrt{2},x\le-\sqrt{2}

Exercise #11

Look at the following function:

4x2410 \frac{\sqrt{4x^2-4}}{10}

What is the domain of the function?

Video Solution

Answer

x1,x1 x\ge1,x\le-1

Exercise #12

Given the following function:

3x2+39 \frac{\sqrt{3x^2+3}}{9}

What is the domain of the function?

Video Solution

Answer

The entire domain

Exercise #13

Look at the following function:

2.5x255 \frac{\sqrt{2.5x^2-5}}{5}

What is the domain of the function?

Video Solution

Answer

x2,x2 x\ge\sqrt{2},x\le-\sqrt{2}

Exercise #14

Look at the following function:

3x2+712 \frac{\sqrt{3x^2+7}}{12}

What is the domain of the function?

Video Solution

Answer

All real numbers

Exercise #15

Look at the following function:

5x2+210 \frac{\sqrt{5x^2+2}}{10}

What is the domain of the function?

Video Solution

Answer

All real numbers

Exercise #16

Look at the following function:

3x2+129 \frac{\sqrt{-3x^2+12}}{9}

What is the domain of the function?

Video Solution

Answer

2x2 -2 \le x \le 2

Exercise #17

Determine the area of the domain without solving the expression:

(4x2)×(7xx6)=2 (\frac{4}{x-2})\times(\frac{7x}{x-6})=2

Video Solution

Answer

x2,x6 x≠2,x≠6

Exercise #18

Find the domain

(no need to resolve)

5x2(x7)=108x \frac{5x}{2(x-7)}=\frac{10}{8x}

Video Solution

Answer

x0,x7 x≠0,x≠7

Exercise #19

Find the area of domain (no need to solve)

x5x6=2x1 \frac{x}{5x-6}=\frac{2}{x-1}

Video Solution

Answer

x1,x115 x≠1,x≠1\frac{1}{5}

Exercise #20

Does the following equation have a true or false value?

x281(x9)(x+9)=1 \frac{x^2-81}{(x-9)(x+9)}=1

Video Solution

Answer

True only when x±9 x\ne\pm9 .