Domain of a Function Practice Problems with Solutions

Master finding domain restrictions for functions with fractions, square roots, and polynomials. Practice exercises with step-by-step solutions and examples.

πŸ“šMaster Domain Restrictions Through Interactive Practice
  • Identify domain restrictions for functions with variables in denominators
  • Find domain limitations for square root functions with algebraic expressions
  • Determine when polynomial functions have unrestricted domains
  • Solve domain problems involving division by zero scenarios
  • Apply algebraic rules to find valid input values for functions
  • Practice with real function examples like f(x) = 1/x and f(x) = √(xΒ²-x-5)

Understanding Indefinite integral

Complete explanation with examples

An integral can be defined for all values (that is, for all X X ). An example of this type of function is the polynomial - which we will study in the coming years.

However, there are integrals that are not defined for all values (all X X ), since if we place certain X X or a certain range of values of X X we will receive an expression considered "invalid" in mathematics. The values of X X for which integration is undefined cause the discontinuity of a function.

integrals that are not defined for all values

Detailed explanation

Practice Indefinite integral

Test your knowledge with 16 quizzes

Look at the following function:

\( \frac{10x-3}{5x-3} \)

What is the domain of the function?

Examples with solutions for Indefinite integral

Step-by-step solutions included
Exercise #1

6x+5=1 \frac{6}{x+5}=1

What is the field of application of the equation?

Step-by-Step Solution

To solve this problem, we will determine the domain, or field of application, of the equation 6x+5=1 \frac{6}{x+5} = 1 .

Step-by-step solution:

  • Step 1: Identify the denominator. In the given equation, the denominator is x+5 x+5 .
  • Step 2: Determine when the denominator is zero. Solve for x x by setting x+5=0 x+5 = 0 .
  • Step 3: Solve the equation: x+5=0 x+5 = 0 gives x=βˆ’5 x = -5 .
  • Step 4: Exclude this value from the domain. The domain is all real numbers except x=βˆ’5 x = -5 .

Therefore, the field of application of the equation is all real numbers except where x=βˆ’5 x = -5 .

Thus, the domain is xβ‰ βˆ’5 x \neq -5 .

Answer:

xβ‰ β‘βˆ’5 x\operatorname{\ne}-5

Video Solution
Exercise #2

x+y:32x+6=4 \frac{x+y:3}{2x+6}=4

What is the field of application of the equation?

Step-by-Step Solution

To solve this problem, we'll follow these steps to find the domain:

  • Step 1: Recognize that the expression x+y:32x+6=4\frac{x+y:3}{2x+6}=4 involves a fraction. The denominator 2x+62x + 6 must not be zero, as division by zero is undefined.
  • Step 2: Set the denominator equal to zero and solve for xx to find the values that must be excluded: 2x+6=02x + 6 = 0.
  • Step 3: Solve 2x+6=02x + 6 = 0:
    • 2x+6=02x + 6 = 0
    • 2x=βˆ’62x = -6
    • x=βˆ’3x = -3
  • Step 4: Conclude that the domain of the function excludes x=βˆ’3x = -3, meaning xβ‰ βˆ’3x \neq -3.

Thus, the domain of the given expression is all real numbers except x=βˆ’3x = -3. This translates to:

xβ‰ β‘βˆ’3 x\operatorname{\ne}-3

Answer:

xβ‰ β‘βˆ’3 x\operatorname{\ne}-3

Video Solution
Exercise #3

3x:4y+6=6 \frac{3x:4}{y+6}=6

What is the field of application of the equation?

Step-by-Step Solution

To determine the field of application of the equation 3x:4y+6=6\frac{3x:4}{y+6}=6, we must identify values of yy for which the equation is defined.

  • The denominator of the given expression is y+6y + 6. In order for the expression to be defined, the denominator cannot be zero.
  • This leads us to solve the equation y+6=0y + 6 = 0.
  • Solving y+6=0y + 6 = 0 gives us y=βˆ’6y = -6.
  • This means y=βˆ’6y = -6 would make the denominator zero, thus the expression would be undefined for this value.

Therefore, the field of application, or the domain of the equation, is all real numbers except y=βˆ’6y = -6.

We must conclude that yβ‰ βˆ’6 y \neq -6 .

Comparing with the provided choices, the correct answer is choice 3: yβ‰ βˆ’6 y \neq -6 .

Answer:

yβ‰ β‘βˆ’6 y\operatorname{\ne}-6

Video Solution
Exercise #4

Given the following function:

5βˆ’x2βˆ’x \frac{5-x}{2-x}

Does the function have a domain? If so, what is it?

Step-by-Step Solution

To determine the domain of the function 5βˆ’x2βˆ’x \frac{5-x}{2-x} , we need to identify and exclude any values of x x that make the function undefined. This occurs when the denominator equals zero.

  • Step 1: Set the denominator equal to zero:
    2βˆ’x=0 2-x = 0
  • Step 2: Solve for x x :
    Adding x x to both sides gives 2=x 2 = x . Hence, x=2 x = 2 .

This means that the function is undefined when x=2 x = 2 . Therefore, the domain of the function consists of all real numbers except x=2 x = 2 .

Thus, the domain is: x≠2 x \ne 2 .

The correct answer choice is:

Yes, x≠2 x\ne2

Answer:

Yes, x≠2 x\ne2

Video Solution
Exercise #5

Given the following function:

49+2xx+4 \frac{49+2x}{x+4}

Does the function have a domain? If so, what is it?

Step-by-Step Solution

To determine the domain of the function 49+2xx+4 \frac{49 + 2x}{x + 4} , we need to focus on avoiding division by zero, which occurs when the denominator is zero.

Let's identify the denominator of the function:

  • The denominator is x+4 x + 4 .

Next, we set the denominator equal to zero and solve for x x :

  • x+4=0 x + 4 = 0
  • Subtract 4 from both sides: x=βˆ’4 x = -4

This calculation shows that the function is undefined when x=βˆ’4 x = -4 . Thus, the domain of the function is all real numbers except x=βˆ’4 x = -4 .

Therefore, the domain of the function is xβ‰ βˆ’4 x \neq -4 .

In terms of the provided choices, this corresponds to choice 4:

Yes, xβ‰ βˆ’4 x \ne -4

Answer:

Yes, xβ‰ βˆ’4 x\ne-4

Video Solution

Frequently Asked Questions

What is the domain of a function and why is it important?

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The domain of a function is the set of all possible input values (x-values) for which the function is defined and produces a valid output. It's important because it tells us which values we can safely substitute into the function without creating mathematical errors like division by zero or negative square roots.

How do you find the domain of a function with a fraction?

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To find the domain of a function with a fraction, identify values that make the denominator equal to zero, then exclude those values from the domain. For example, with f(x) = 1/x, set the denominator x β‰  0, so the domain is all real numbers except x = 0.

What are the domain restrictions for square root functions?

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For square root functions, the expression under the square root must be greater than or equal to zero (non-negative). Set the expression β‰₯ 0 and solve the inequality to find the domain. For f(x) = √(xΒ²-x-5), solve xΒ²-x-5 β‰₯ 0 to determine valid x-values.

Do polynomial functions have domain restrictions?

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No, polynomial functions have no domain restrictions and are defined for all real numbers. This is because polynomials only involve addition, subtraction, and multiplication operations, which never create undefined expressions like division by zero or negative square roots.

What causes a function to be undefined at certain points?

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Functions become undefined when mathematical operations create invalid expressions. The most common causes are: 1) Division by zero (denominators equal zero), 2) Even roots of negative numbers (like √(-4)), 3) Logarithms of non-positive numbers, and 4) Inverse trigonometric functions outside their restricted ranges.

How do you write domain using interval notation?

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Domain is written using interval notation with parentheses () for excluded values and brackets [] for included values. For example: (-∞, 0) βˆͺ (0, ∞) means all real numbers except 0, while [2, ∞) means all numbers greater than or equal to 2.

What is the difference between domain and range of a function?

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Domain refers to all possible input values (x-values) that can be used in a function, while range refers to all possible output values (y-values) that the function can produce. Domain restrictions are determined by the function's mathematical limitations, while range depends on the function's behavior and graph.

How do you solve domain problems with complex expressions?

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For complex expressions, identify all restrictions separately then combine them: 1) Find where denominators equal zero, 2) Determine where expressions under even roots are negative, 3) Check for other undefined operations, 4) Combine all restrictions to find the final domain, excluding all problematic values.

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