Indefinite integral - Examples, Exercises and Solutions

Since the function's denominator equals 4, the domain of the function is all real numbers. This means that any one of the x values could be compatible.

Since the unknown is in the denominator, we should remember that the denominator cannot be equal to 0.

In other words, $x\ne0$

The domain of the function is all those values that, when substituted into the function, will make the function legal and defined.

The domain in this case will be all real numbers that are not equal to 0.

The denominator of the function cannot be equal to 0.

Therefore, we will set the denominator equal to 0 and solve for the domain:

$(2x-2)^2\ne0$

$2x\ne2$

$x\ne1$

In other words, the domain of the function is all numbers except 1.

Since the denominator is positive for all $x$, the domain of the function is the entire domain.

That is, all values of $x$. Therefore, there is no domain limits.

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 multiplication between fractions whose denominators contain algebraic expressions that include the variable of the equation (which we are looking for when solving the equation),

Of course, these fractions are 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, meaning:

For the fraction in parentheses in the expression on the left side we get:

$\boxed{ x\neq0}$

For the fraction in parentheses in the expression on the right side we get:

$x+1\neq0 \\$Let's 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 of both sides by a negative number requires reversing the trend.

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 multiplication between fractions whose denominators contain algebraic expressions that include the variable of the equation (which we are looking for when solving the equation),

Of course, these fractions are defined as long as the expression in their denominators is 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, meaning:

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

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

$13x\neq0$

$$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.