To find the critical points of the function $f(x) = \frac{-x^2 - 1}{x^2 - 4}$, we need to determine where the derivative $f'(x)$ is equal to zero or undefined.

Step 1: Find the derivative $f'(x)$

We'll use the quotient rule, which states: $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$ where $g(x) = -x^2 - 1$ and $h(x) = x^2 - 4$.

1. $g(x) = -x^2 - 1$, so $g'(x) = -2x$.
2. $h(x) = x^2 - 4$, so $h'(x) = 2x$.

Now, applying the quotient rule: $f'(x) = \frac{(-2x)(x^2 - 4) - (-x^2 - 1)(2x)}{(x^2 - 4)^2}$

Step 2: Simplify the expression

Expand and simplify the numerator: $f'(x) = \frac{-2x(x^2 - 4) + 2x(x^2 + 1)}{(x^2 - 4)^2}$ $f'(x) = \frac{-2x^3 + 8x + 2x^3 + 2x}{(x^2 - 4)^2}$ $f'(x) = \frac{10x}{(x^2 - 4)^2}$

Step 3: Set the derivative equal to zero to find critical points

For $f'(x) = 0$: $\frac{10x}{(x^2 - 4)^2} = 0$ This equation equals zero when the numerator is zero: $10x = 0 \quad \Rightarrow \quad x = 0$

Step 4: Verify whether the denominator affects critical points

The function is undefined where the denominator $(x^2 - 4)^2$ is zero, which occurs at: $x^2 - 4 = 0 \quad \Rightarrow \quad x = \pm 2$ These points are not in the domain of the function, so they are not considered critical points.

Conclusion:

The function $f(x)$ has a critical point at $x = 0$.

Would you like more details or have any questions?

Related Questions:

1. What are the second derivative test conditions for classifying the critical point at $x = 0$?
2. How would the graph of $f(x)$ behave near the critical points?
3. What is the behavior of $f(x)$ as $x$ approaches $\pm 2$?
4. How do asymptotes influence the analysis of critical points in rational functions?
5. What other methods can be used to find critical points besides differentiation?

Tip:

Always consider the domain of the function when finding critical points, as some potential points might lie outside the domain and thus aren't valid.