Exercises on the Domain of a Function

The domain of an entire polynomial function is , that is, all real numbers.

 

1

 

 

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3

 

Note that this is a polynomial function with rational coefficients:


Therefore,

 

4

 

Note that this is a polynomial function with rational coefficients, therefore,

 

5

 

Note that this is a polynomial function with rational and irrational coefficients, therefore,

The domain of a rational function is minus the values that make the denominator zero. To find the domain, we must set the denominator equal to zero and solve the equation. The solutions to that equation are the points that do not belong to the domain, since they make the denominator zero.

 

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because this equation has no real roots.

 

7

 

 

because the root is double.

 

8

 

We have that , then

 

 

Thus, or . Therefore, the domain is

 

9

We observe that the polynomial is the expansion of a binomial cubed

because is a triple root.

 

10

 

Factoring

 

Therefore,

The domain of an irrational function with odd index is

 

The domain of an irrational function with even index is obtained from the points that satisfy that the radicand is greater than or equal to zero.

 

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The domain of this function is all reals minus the values where the denominator of the rational function inside the cube root becomes zero. Thus,

 

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The domain of this function is all reals minus the values where the denominator of the rational function inside the cube root becomes zero. Thus,

 

 

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We solve the inequality

 

 

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We solve

 

9

 

We solve
We set equal to zero to obtain the roots

 

 

Finally, we take the intervals where the inequality is positive. The union of these will be our domain. Therefore

 

10

 

We solve
We set equal to zero to obtain the roots

 

 

Finally, we take the intervals where the inequality is non-positive. The union of these will be our domain. Therefore

 

11

 

We solve because is always greater than or equal to zero.

 

12

 

We solve

If we set equal to zero, the corresponding equation has no real solutions. Furthermore, we note that for any value taken it will be positive or zero. Therefore,

 

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We solve

Note that this inequality is only satisfied for the value since for all other values of , the result is always negative. Thus, is the only value that satisfies our inequality and therefore,

 

14

 

We solve the quadratic inequality

 

 

Finally, we take the intervals where the inequality is positive with the endpoints where it equals zero included. The union of these intervals will be our domain. Therefore,

 

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We solve

 

 

Finally, we take the intervals where the inequality is positive with the endpoints where it equals zero included. The union of these intervals will be our domain. Therefore,

 

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Since the root is in the denominator, the radicand must be greater than zero, but not equal because then it would make the denominator zero.

We solve

 

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In this case, the denominator must be non-zero and the root in the numerator must be greater than or equal to zero.

We solve

The solution is the intersection of the two sets, therefore,

 

18

 

The numerator must be greater than or equal to zero and the denominator must be non-zero.

We solve

 

 

The solution is the intersection of the two sets, therefore,

 

19

 

The numerator must be greater than or equal to zero and the denominator must be non-zero.

We solve

 

 

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The radicand must be greater than or equal to zero and the denominator must be non-zero.

We solve y

 

 

Using sign analysis with critical points at , we get

The domain of an exponential function is

 

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Since the exponent is rational, does not belong to the domain because it makes the denominator zero. Therefore, .

For the logarithm function to be well defined, its argument must be positive, that is, its domain is

 

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We solve

 

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Since is always positive for , then

 

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Since is always positive, then

 

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We solve

 

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Since the denominator is always positive, we only study the numerator. Thus

 

 

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The domain of an irrational function with an even index is formed by the set of values that make the radicand greater than or equal to zero. The trigonometric functions sine and cosine have all real numbers as their domain. Additionally, their maximum value is 1, so these functions always have values less than or equal to 1 for every real number.

 

 

4

 

We solve

 

5

 

We solve

1

Since each piece is polynomial, then

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Since each piece is polynomial, then

3

Since each piece is polynomial, then

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Since each piece is polynomial, then

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In the first piece, the denominator must be non-zero. In the second piece, since 3 is a constant it will always be positive, we only need to study that the denominator is greater than zero. Thus,

Finally, the solution is