Math Problem Statement

solve

Find the domain of each function.

Solution

Let's go through each function in the image and find its domain:

  1. y=4x+3y = 4x + 3 (Part a):

    • This is a linear function, and linear functions are defined for all real numbers.
    • Domain: (,)(-\infty, \infty)
  2. y=7xy = \frac{7}{x} (Part b):

    • This is a rational function. The only restriction is that x0x \neq 0 because division by zero is undefined.
    • Domain: (,0)(0,)(-\infty, 0) \cup (0, \infty)
  3. y=1x5y = \frac{1}{x - 5} (Part c):

    • Another rational function. The restriction is that x5x \neq 5, since division by zero is undefined.
    • Domain: (,5)(5,)(-\infty, 5) \cup (5, \infty)
  4. y=3x+8y = \frac{3}{x + 8} (Part d):

    • This is a rational function with the restriction x8x \neq -8 to avoid division by zero.
    • Domain: (,8)(8,)(-\infty, -8) \cup (-8, \infty)
  5. y=x2y = \sqrt{x - 2} (Part e):

    • This is a square root function, and the expression inside the square root must be non-negative. So, x20x - 2 \geq 0, meaning x2x \geq 2.
    • Domain: [2,)[2, \infty)
  6. y=2x2+3y = 2x^2 + 3 (Part f):

    • This is a quadratic function, which is defined for all real numbers.
    • Domain: (,)(-\infty, \infty)
  7. y=2x+5y = \frac{2}{x + 5} (Part g):

    • This is a rational function with the restriction x5x \neq -5 to avoid division by zero.
    • Domain: (,5)(5,)(-\infty, -5) \cup (-5, \infty)
  8. y=x+6y = \sqrt{x + 6} (Part h):

    • This is a square root function. The expression inside the square root must be non-negative, so x+60x + 6 \geq 0, meaning x6x \geq -6.
    • Domain: [6,)[-6, \infty)

Summary of Domains:

  • (a) (,)(-\infty, \infty)
  • (b) (,0)(0,)(-\infty, 0) \cup (0, \infty)
  • (c) (,5)(5,)(-\infty, 5) \cup (5, \infty)
  • (d) (,8)(8,)(-\infty, -8) \cup (-8, \infty)
  • (e) [2,)[2, \infty)
  • (f) (,)(-\infty, \infty)
  • (g) (,5)(5,)(-\infty, -5) \cup (-5, \infty)
  • (h) [6,)[-6, \infty)

Would you like more details on any of these functions?


Here are 5 follow-up questions:

  1. How do you find the domain of a rational function?
  2. What happens to the domain when a square root function is involved?
  3. Can the domain of a linear function ever have restrictions?
  4. How would you graph the domain of y=1x5y = \frac{1}{x-5}?
  5. What are some real-world applications of domain restrictions in functions?

Tip: For rational functions, always look out for values of xx that make the denominator zero to avoid undefined values.

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Math Problem Analysis

Mathematical Concepts

Domain of Functions
Rational Functions
Square Root Functions

Formulas

Domain of a function: Restrictions on x-values for which the function is defined
Rational function: y = 1/(x - a) has restrictions when x = a
Square root function: y = √(x - a) requires x - a ≥ 0

Theorems

A function is undefined where its denominator equals zero (for rational functions).
Square roots are defined for non-negative numbers.

Suitable Grade Level

Grades 9-12