Key Concepts and Formulas

• First Derivative Test: A method to find local extrema by analyzing the sign changes of the first derivative around critical points.
• Critical Points: Points where the first derivative is zero or undefined.
• Local Minimum: Occurs at a critical point where the first derivative changes from negative to positive.
• Local Maximum: Occurs at a critical point where the first derivative changes from positive to negative.

Step-by-Step Solution

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

We are given the function $f(x) = 9x^4 + 12x^3 - 36x^2 + 25$. To find the critical points, we first need to calculate the first derivative.

$f'(x) = \frac{d}{dx}(9x^4 + 12x^3 - 36x^2 + 25)$ Applying the power rule and sum/difference rule: $f'(x) = 36x^3 + 36x^2 - 72x$

Step 2: Find the critical points.

Critical points occur where $f'(x) = 0$.

$36x^3 + 36x^2 - 72x = 0$ Factor out the common factor $36x$: $36x(x^2 + x - 2) = 0$ Factor the quadratic: $36x(x + 2)(x - 1) = 0$ The critical points are: $x = 0, x = -2, x = 1$

Step 3: Apply the First Derivative Test.

We create a sign chart to analyze the sign of $f'(x)$ in the intervals defined by the critical points. This will help us determine where the function is increasing or decreasing. The intervals are $(-\infty, -2)$, $(-2, 0)$, $(0, 1)$, and $(1, \infty)$.

From the sign chart:

• $f'(x)$ changes from negative to positive at $x = -2$, so there's a local minimum at $x = -2$.
• $f'(x)$ changes from positive to negative at $x = 0$, so there's a local maximum at $x = 0$.
• $f'(x)$ changes from negative to positive at $x = 1$, so there's a local minimum at $x = 1$.

Therefore, the set of local minimum points is $S_1 = \{-2, 1\}$ and the set of local maximum points is $S_2 = \{0\}$.

Step 4: Check against options.

The correct answer should be S 1 = {–2, 1}; S 2 = {0} which corresponds to option (D). However, the provided "Correct Answer" is option (A), which has been corrected to the correct option (D) based on the calculated local minima and maxima. The provided answer key is wrong.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when evaluating $f'(x)$ in each interval. A single sign error will lead to the wrong conclusion about increasing/decreasing behavior and thus about the nature of the extrema.
• Factoring: Always factor $f'(x)$ completely to easily identify the roots and their corresponding intervals.
• Test Values: Choose test values that are strictly within the intervals and easy to evaluate in $f'(x)$.

Summary

We found the local minimum and maximum points of the given function by first finding its derivative, then identifying the critical points by setting the derivative equal to zero. We then used the first derivative test (sign chart) to determine the intervals where the function is increasing or decreasing, allowing us to classify each critical point as either a local minimum or local maximum. This gave us the sets $S_1 = \{-2, 1\}$ and $S_2 = \{0\}$.

The final answer is \boxed{D}.