We are given the derivative of the function $f'(x) = (x-a)(x+a) = x^2 - a^2$, where $a$ is a positive constant. We are tasked with analyzing the function's behavior based on the given options.

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

The derivative $f'(x) = (x-a)(x+a) = x^2 - a^2$ is a quadratic function. To understand the sign of $f'(x)$, we need to analyze when $f'(x)$ is positive, negative, or zero.

• Critical points: The function $f'(x) = 0$ at $x = \pm a$. These are the points where the derivative changes sign.

Step 2: Determine the sign of $f'(x)$ in different intervals

We can check the sign of $f'(x)$ in the intervals defined by the critical points $x = -a$ and $x = a$. The intervals are:

1. $x < -a$
2. $-a < x < a$
3. $x > a$

• For $x < -a$: In this region, both $x-a$ and $x+a$ are negative, so $f'(x) = (x-a)(x+a) > 0$.
• For $-a < x < a$: In this region, $x-a$ is negative and $x+a$ is positive, so $f'(x) = (x-a)(x+a) < 0$.
• For $x > a$: In this region, both $x-a$ and $x+a$ are positive, so $f'(x) = (x-a)(x+a) > 0$.

Step 3: Interpretation of the statements

• Statement 1: "f is decreasing for $-a < x < a$ because $f'(x) < 0$ for $-a < x < a$."

  • True: As shown above, $f'(x) < 0$ for $-a < x < a$, so the function is decreasing in this interval.

• Statement 2: "f is decreasing for $x < -a$ and $x > a$ because $f'(x) < 0$ for $x < -a$ and $x > a$."

  • False: In both intervals $x < -a$ and $x > a$, $f'(x) > 0$, meaning the function is increasing, not decreasing.

• Statement 3: "f is decreasing for $x < 0$ because $f'(x) < 0$ for $x < 0$."

  • False: For $x < 0$, $f'(x) = (x-a)(x+a)$. This expression is positive when $x < -a$ and negative when $-a < x < 0$. So, the function is not decreasing for the entire interval $x < 0$.

• Statement 4: "f is decreasing for $x < 0$ because $f''(x) < 0$ for $x < 0$."

  • False: This is a misunderstanding. While $f'(x)$ is negative for $-a < x < a$, the second derivative $f''(x)$ is not necessarily negative in this region. In fact, calculating $f''(x)$ and analyzing it would give a different insight into concavity rather than monotonicity.

Conclusion

The true statement is:

• f is decreasing for $-a < x < a$ because $f'(x) < 0$ for $-a < x < a$.

Would you like further explanation on any of these points or need help with another topic?

Follow-up Questions:

1. What does the second derivative $f''(x)$ tell us about the concavity of the function?
2. Can you identify the critical points of a function using its derivative?
3. How does the sign of $f'(x)$ relate to intervals of increase or decrease for a function?
4. How can you determine whether a function is concave up or concave down from its second derivative?
5. What is the relationship between the points where $f'(x) = 0$ and the function's behavior?

Tip: Remember that when analyzing a function's behavior, always check the sign of the derivative in different intervals, as it directly indicates whether the function is increasing or decreasing in that region!