Math Problem Statement

Determine the domain of the following functions: a) f(x) = sqrt(x + 3) b) g(x) = 3 / (x^2 - 4) c) h(x) = x^2 + 7x - 3 d) p(x) = sqrt(7 - 4x) e) k(x) = sqrt(x - 2) / (x - 9) f) n(x) = x / (x + 2) g) j(x) = x / 9 h) m(x) = 9 / x

Solution

Let's go through each function one by one to determine its domain.

8a. f(x)=x+3f(x) = \sqrt{x + 3}

The square root function y\sqrt{y} is defined only for non-negative values of yy. Thus, for x+3\sqrt{x + 3} to be defined, we need:

x+30x + 3 \geq 0 x3x \geq -3

Domain: x3x \geq -3, or [3,)[-3, \infty).


8b. g(x)=3x24g(x) = \frac{3}{x^2 - 4}

The function ab\frac{a}{b} is undefined when the denominator b=0b = 0. So, we need to find when x24=0x^2 - 4 = 0:

x24=(x2)(x+2)=0x^2 - 4 = (x - 2)(x + 2) = 0 x=2orx=2x = 2 \quad \text{or} \quad x = -2

Thus, the function is undefined at x=2x = 2 and x=2x = -2.

Domain: x(,2)(2,2)(2,)x \in (-\infty, -2) \cup (-2, 2) \cup (2, \infty).


8c. h(x)=x2+7x3h(x) = x^2 + 7x - 3

This is a quadratic function, which is defined for all real values of xx. Therefore, there are no restrictions on the domain.

Domain: (,)(-\infty, \infty).


8d. p(x)=74xp(x) = \sqrt{7 - 4x}

For the square root to be defined, we need the expression inside the square root to be non-negative:

74x07 - 4x \geq 0 4x74x \leq 7 x74x \leq \frac{7}{4}

Domain: (,74](-\infty, \frac{7}{4}].


8e. k(x)=x2x9k(x) = \frac{\sqrt{x - 2}}{x - 9}

  1. The square root in the numerator x2\sqrt{x - 2} requires: x20x2x - 2 \geq 0 \quad \Rightarrow \quad x \geq 2

  2. The denominator x90x - 9 \neq 0, so x9x \neq 9.

Thus, the domain is x2x \geq 2 but x9x \neq 9.

Domain: [2,9)(9,)[2, 9) \cup (9, \infty).


8f. n(x)=xx+2n(x) = \frac{x}{x + 2}

The function is undefined when the denominator is zero, so x+20x + 2 \neq 0:

x2x \neq -2

Domain: x(,2)(2,)x \in (-\infty, -2) \cup (-2, \infty).


8g. j(x)=x9j(x) = \frac{x}{9}

This is a linear function, which is defined for all real values of xx.

Domain: (,)(-\infty, \infty).


8h. m(x)=9xm(x) = \frac{9}{x}

The function is undefined when the denominator is zero, so x0x \neq 0.

Domain: x(,0)(0,)x \in (-\infty, 0) \cup (0, \infty).


Do you need more details for any of the solutions?

Here are some relative questions to extend your understanding:

  1. How do we determine the domain for a function involving a square root?
  2. Why do rational functions have undefined values at certain points?
  3. What makes quadratic functions have domains over all real numbers?
  4. How can we apply graphing to visualize the domain of these functions?
  5. How does solving inequalities help in determining the domain?

Tip: Always check for restrictions like square roots and division by zero when determining the domain of a function.

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Math Problem Analysis

Mathematical Concepts

Domain of Functions
Square Roots
Rational Functions
Quadratic Equations

Formulas

sqrt(y) is defined for y >= 0
Rational functions are undefined when the denominator equals zero

Theorems

Square Root Rule: sqrt(x) is defined when x >= 0
Rational Function Undefined Rule: Functions are undefined when their denominator is zero

Suitable Grade Level

Grades 9-12