Math Problem Statement

Tentukan domain dan range dari: a. f(x) = √(2x + 3), b. g(x) = 1 / (4x - 1), c. h(x) = √((x + 1)^-1)

Solution

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The question asks to find the domain and range for three functions:

  1. f(x)=2x+3f(x) = \sqrt{2x + 3}
  2. g(x)=14x1g(x) = \frac{1}{4x - 1}
  3. h(x)=(x+1)1h(x) = \sqrt{(x + 1)^{-1}}

Solutions:

  1. For f(x)=2x+3f(x) = \sqrt{2x + 3}:

    • Domain: Since square roots require the argument to be non-negative, we need 2x+302x + 3 \geq 0. Solving this gives x32x \geq -\frac{3}{2}. Therefore, the domain is: x[32,)x \in \left[ -\frac{3}{2}, \infty \right)
    • Range: The output of a square root function is non-negative, so the range is: f(x)0orf(x)[0,)f(x) \geq 0 \quad \text{or} \quad f(x) \in [0, \infty)
  2. For g(x)=14x1g(x) = \frac{1}{4x - 1}:

    • Domain: The function is undefined when the denominator is zero, i.e., 4x1=04x - 1 = 0 implies x=14x = \frac{1}{4}. Thus, the domain is all real numbers except x=14x = \frac{1}{4}: x(,14)(14,)x \in (-\infty, \frac{1}{4}) \cup (\frac{1}{4}, \infty)
    • Range: Since this is a rational function, g(x)g(x) can take any real value except 0, which is asymptotically approached but never reached. Hence, the range is: g(x)(,0)(0,)g(x) \in (-\infty, 0) \cup (0, \infty)
  3. For h(x)=(x+1)1h(x) = \sqrt{(x + 1)^{-1}}:

    • Domain: The argument of the square root must be non-negative, i.e., (x+1)10(x + 1)^{-1} \geq 0. This implies that x+1>0x + 1 > 0, so x>1x > -1. Therefore, the domain is: x(1,)x \in (-1, \infty)
    • Range: Since the square root function is non-negative, the range of this function is: h(x)[0,)h(x) \in [0, \infty)

Would you like more details or clarifications? Here are five related questions to further explore these concepts:

  1. How do we determine the domain of a function involving a square root?
  2. What does the behavior of rational functions near vertical asymptotes tell us about their range?
  3. How would the domain of a function change if the expression under the square root was negative?
  4. Can you explain why g(x)g(x) never equals zero?
  5. How do the properties of inverse powers, like in h(x)=(x+1)1h(x) = \sqrt{(x+1)^{-1}}, influence the function's domain?

Tip: When finding the domain of functions involving square roots, always start by solving the inequality inside the square root for non-negative values.

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Math Problem Analysis

Mathematical Concepts

Domain and Range
Square Root Functions
Rational Functions
Inequalities

Formulas

f(x) = √(2x + 3)
g(x) = 1 / (4x - 1)
h(x) = √((x + 1)^-1)

Theorems

The domain of a square root function requires the argument to be non-negative.
The domain of a rational function excludes values where the denominator is zero.

Suitable Grade Level

Grades 11-12