Math Problem Statement

Determine the domain and range of the function f(x) = 1 / (2x^2 - x - 6).

Solution

The problem in the image remains the same:

f(x)=12x2x6f(x) = \frac{1}{2x^2 - x - 6}

The task is to determine the domain and range of this function. Let's summarize and verify the steps for each.


Domain of f(x)f(x)

To find the domain, we need to identify values of xx that make the denominator zero, as these are the points where f(x)f(x) is undefined. We start by setting the denominator equal to zero and solving for xx:

2x2x6=02x^2 - x - 6 = 0

Solving the Quadratic Equation

Using the quadratic formula x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=2a = 2, b=1b = -1, and c=6c = -6:

  1. Substitute the values: x=(1)±(1)242(6)22x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-6)}}{2 \cdot 2}

  2. Simplify under the square root: x=1±1+484=1±494x = \frac{1 \pm \sqrt{1 + 48}}{4} = \frac{1 \pm \sqrt{49}}{4}

  3. Since 49=7\sqrt{49} = 7, we get: x=1+74=2andx=174=32x = \frac{1 + 7}{4} = 2 \quad \text{and} \quad x = \frac{1 - 7}{4} = -\frac{3}{2}

Therefore, the function is undefined at x=2x = 2 and x=32x = -\frac{3}{2}.

Domain Conclusion

The domain of f(x)f(x) is all real numbers except x=2x = 2 and x=32x = -\frac{3}{2}. In interval notation:

Domain of f(x)=(,32)(32,2)(2,)\text{Domain of } f(x) = (-\infty, -\frac{3}{2}) \cup \left(-\frac{3}{2}, 2\right) \cup (2, \infty)


Range of f(x)f(x)

To find the range, we examine the output values of f(x)f(x), particularly by looking at its behavior at infinity and near the points where it is undefined.

Behavior as x±x \to \pm \infty

  • As xx becomes very large in either the positive or negative direction, the 2x22x^2 term in the denominator dominates, making f(x)12x2f(x) \approx \frac{1}{2x^2}, which approaches 0.
  • Thus, f(x)f(x) gets closer to zero but never actually reaches it. Therefore, y=0y = 0 is a horizontal asymptote.

Behavior Near the Vertical Asymptotes

  • At x=2x = 2 and x=32x = -\frac{3}{2}: The function goes to ±\pm \infty near these values due to division by zero, indicating vertical asymptotes at these points.
  • This behavior allows f(x)f(x) to take on both very large positive and negative values, suggesting that the function can reach any value except zero.

Range Conclusion

Since f(x)f(x) approaches all real values except zero, the range of f(x)f(x) is:

Range of f(x)=(,0)(0,)\text{Range of } f(x) = (-\infty, 0) \cup (0, \infty)


Final Answer

  • Domain: (,32)(32,2)(2,)(-\infty, -\frac{3}{2}) \cup \left(-\frac{3}{2}, 2\right) \cup (2, \infty)
  • Range: (,0)(0,)(-\infty, 0) \cup (0, \infty)

Would you like additional clarification on any part, or help with similar problems?


Here are five questions that further explore related concepts:

  1. How would the domain change if the function were f(x)=1x2x6f(x) = \frac{1}{x^2 - x - 6}?
  2. What is the significance of horizontal and vertical asymptotes in determining the range?
  3. Can a rational function have a range that includes zero?
  4. How does factoring help in determining the points where a rational function is undefined?
  5. How do transformations affect the asymptotes of rational functions?

Tip: When dealing with rational functions, always identify where the denominator is zero to determine the domain, then examine limits to understand the range.

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Math Problem Analysis

Mathematical Concepts

Algebra
Rational Functions
Domain and Range
Quadratic Equations
Asymptotes

Formulas

Quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / (2a)

Theorems

Properties of rational functions and asymptotes

Suitable Grade Level

Grades 10-12