To solve this problem, we will use the second-derivative test on the function $y = x^3 - 12x + 3$.

Step 1: Find the first derivative

The first derivative $y'$ gives us the critical points, where relative maxima or minima might occur:

$y' = \frac{d}{dx}(x^3 - 12x + 3) = 3x^2 - 12$

Step 2: Set the first derivative equal to zero

To find the critical points, we solve $y' = 0$:

$3x^2 - 12 = 0$

$x^2 = 4$

$x = \pm 2$

So, the critical points are $x = 2$ and $x = -2$.

Step 3: Find the second derivative

The second derivative $y''$ helps us determine the concavity at these critical points:

$y'' = \frac{d}{dx}(3x^2 - 12) = 6x$

Step 4: Apply the second-derivative test

• At $x = 2$, we compute $y''(2)$: $y''(2) = 6(2) = 12$ Since $y''(2) > 0$, the function is concave up at $x = 2$, meaning this is a relative minimum.

• At $x = -2$, we compute $y''(-2)$: $y''(-2) = 6(-2) = -12$ Since $y''(-2) < 0$, the function is concave down at $x = -2$, meaning this is a relative maximum.

Step 5: Conclusion

• Relative maximum occurs at $x = -2$.
• Relative minimum occurs at $x = 2$.

Thus, the correct answer is:

Choice A: The relative maxima occur at $x = -2$. The relative minima occur at $x = 2$.

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Would you like more details or have any questions?

Here are 5 related questions to consider:

1. What is the absolute maximum and minimum of this function?
2. How do you use the first-derivative test to confirm relative extrema?
3. Can you find the inflection points of the function $y = x^3 - 12x + 3$?
4. What is the behavior of the function as $x \to \infty$ and $x \to -\infty$?
5. How do the first and second derivatives relate to the graph of the function?

Tip: Always check the concavity at critical points using the second derivative for better insight into the function's behavior.