To find the critical points of the function $f(x) = \ln(x^2 - 4)$, we need to follow these steps:

Step 1: Determine the Domain

The function $f(x) = \ln(x^2 - 4)$ is only defined when $x^2 - 4 > 0$. This inequality implies: $x^2 > 4 \Rightarrow x > 2 \text{ or } x < -2$ So, the domain of $f(x)$ is $x > 2$ or $x < -2$.

Step 2: Find $f'(x)$

To find the critical points, we first need the derivative $f'(x)$ and set it equal to zero. Using the chain rule: $f(x) = \ln(x^2 - 4)$ $f'(x) = \frac{1}{x^2 - 4} \cdot 2x = \frac{2x}{x^2 - 4}$

Step 3: Set $f'(x) = 0$

Setting $f'(x) = 0$: $\frac{2x}{x^2 - 4} = 0$ This equation will be zero when the numerator is zero, i.e., $2x = 0$, which implies $x = 0$.

Step 4: Check if $x = 0$ is in the Domain

Since $x = 0$ does not satisfy the domain $x > 2$ or $x < -2$, it is not a valid critical point for this function.

Conclusion

There are no values of $x$ within the domain that make $f'(x) = 0$. Therefore, there are no critical points.

The correct answer is:

• (a) zero

Would you like more details or have any questions?

Here are some related questions to expand understanding:

1. What would the critical points be if the domain were extended?
2. How would we approach finding critical points for a function with a different logarithmic argument?
3. What role does the domain of a function play in determining critical points?
4. How does the chain rule help in differentiating functions of functions?
5. What are the implications of a function having no critical points on its graph?

Tip: Always check the domain first when dealing with functions involving logarithms or square roots, as it often influences the existence of critical points.