Domain of a Function - 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 domain depends on the denominator and we can see that there is no variable in the denominator.

Therefore, the domain is all numbers.

Let's examine the given expression:

$\frac{x}{16}$

As we know, the only restriction that applies to division operation is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

However in the given expression:

$\frac{x}{16}$

the denominator is 16 and:

$16\neq0$

Therefore the fraction is well defined and thus the unknown, which is in the numerator, can take any value,

Meaning - the domain (definition range) of the given expression is:

all x

(This means that we can substitute any number for the unknown x and the expression will remain well defined),

Therefore the correct answer is answer B.

The domain of a fraction depends on the denominator.

Since you cannot divide by zero, the denominator of a fraction cannot equal zero.

Therefore, for the fraction $$$\frac{6}{x}$, the domain is "All numbers except 0," since the denominator cannot equal zero.

In other words, the domain is:

$x\ne0$

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.

Let's examine the given expression:

$\frac{x+8}{3x}$

$$As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

$\frac{x+8}{3x}$

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

$$$3x\neq0$

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

$3x\neq0\hspace{6pt}\text{/}:3 \\ \boxed{x\neq 0}$

Therefore, the domain (definition domain) of the given expression is:

$x\neq 0$

(This means that if we substitute any number different from $0$ for x, the expression will remain well-defined),

Therefore, the correct answer is answer A.

Note:

In a general form - solving an inequality of this form, meaning, a non-linear, but point inequality - which uses the $\neq$ sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every aspect to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Let's examine the given expression:

$\frac{3}{x+2}$

$$As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

$\frac{3}{x+2}$

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

$$$x+2\neq0$

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

$x+2\neq0 \\ \boxed{x\neq -2}$

Therefore, the domain (definition domain) of the given expression is:

$x\neq -2$

(This means that if we substitute for the variable x any number different from$(-2)$the expression will remain well-defined),

Therefore, the correct answer is answer D.

Note:

In a general way - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the $\neq$ sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every aspect to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Let's examine the given expression:

$\frac{7}{13+x}$

$$As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

$\frac{7}{13+x}$

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

$$$13+x\neq0$

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

$13+x\neq0 \\ \boxed{x\neq -13}$

Therefore, the domain (definition domain) of the given expression is:

$x\neq -13$

(This means that if we substitute any number different from $(-13)$ for the variable x, the expression will remain well-defined),

Therefore, the correct answer is answer D.

Note:

In a general way - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the $\neq$ sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Let's examine the given expression:

$\frac{8}{-2+x}$

$$As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

$\frac{8}{-2+x}$

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

$$$-2+x\neq0$

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

$-2+x\neq0 \\ \boxed{x\neq 2}$

Therefore, the domain (definition domain) of the given expression is:

$x\neq 2$

(This means that if we substitute any number different from $2$ for x, the expression will remain well-defined),

Therefore, the correct answer is answer C.

Note:

In a general form - solving an inequality of this form, meaning, a non-graphical, but point inequality - that uses the $\neq$ sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

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.

Let's examine the given expression:

$\frac{-8-x}{-x}$

$$As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

$\frac{-8-x}{-x}$

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

$$$-x\neq0$

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

$-x\neq0 \hspace{6pt}\text{/}:(-1) \\ \boxed{x\neq 0}$

Therefore, the domain (definition domain) of the given expression is:

$x\neq 0$

(This means that if we substitute for the variable x any number different from$0$the expression will remain well-defined),

Therefore, the correct answer is answer A.

Note:

In a general form - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the $\neq$ sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Let's examine the given expression:

$\frac{x}{x+3}$

$$As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

$\frac{x}{x+3}$

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

$$$x+3\neq0$

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

$x+3\neq0 \\ \boxed{x\neq -3}$

Therefore, the domain (definition domain) of the given expression is:

$x\neq -3$

(This means that if we substitute for the variable x any number different from$(-3)$the expression will remain well-defined),

Therefore, the correct answer is answer D.

Note:

In a general form - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the $\neq$ sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Let's examine the given expression:

$\frac{-8-x}{-3x+2}$

$$As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

$\frac{-8-x}{-3x+2}$

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

$$$-3x+2\neq0$

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

$-3x+2\neq0 \\ -3x\neq-2\hspace{6pt}\text{/}:(-3)\\ x\neq\frac{-2}{-3}\\ \boxed{x\neq \frac{2}{3}}$

Therefore, the domain (definition domain) of the given expression is:

$x\neq \frac{2}{3}$

(This means that if we substitute any number different from $\frac{2}{3}$ for x, the expression will remain well-defined),

Therefore, the correct answer is answer C.

Note:

In a general form - solving an inequality of this form, meaning, a non-graphical, but point inequality - that uses the $\neq$ sign and not the slope signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in an identical way and all rules used to solve an equation of any type are identical for it as well.