The problem in the image remains the same:

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

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

Domain of $f(x)$

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

$2x^2 - x - 6 = 0$

Solving the Quadratic Equation

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

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

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

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

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

Domain Conclusion

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

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

Range of $f(x)$

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

Behavior as $x \to \pm \infty$

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

Behavior Near the Vertical Asymptotes

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

Range Conclusion

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

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

Final Answer

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

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 $f(x) = \frac{1}{x^2 - x - 6}$?
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.