Test for relative maxima and minima. Use the​ second-derivative test, if possible.
 yequalsxcubedminus12xplus3
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Part 1
Select the correct choice below​ and, if​ necessary, fill in the answer​ box(es) to complete your choice.
A.
The relative maxima occur at xequals
  
enter your response here. The relative minima occur at 
  
enter your response here.
​(Type integers or simplified fractions. Use a comma to separate answers as​ needed.)
B.
The relative maxima occur at xequals
  
enter your response here. There are no relative minima.
​(Type an integer or a simplified fraction. Use a comma to separate answers as​ needed.)
C.
The relative minima occur at xequals
  
enter your response here. There are no relative maxima.
​(Type an integer or a simplified fraction. Use a comma to separate answers as​ needed.)
D.
There are no relative maxima and no relative minima.

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 \( x = 2 \).

---

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.

Test for relative maxima and minima. Use the second-derivative test, if possible.
 y = x^3 - 12x + 3
Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

This detailed solution walks you through finding the relative maxima and minima for the function y = x^3 - 12x + 3 using the second derivative test. Learn how to calculate critical points, apply the second-derivative test, and identify whether the function has relative maxima or minima. Suitable for high school students in Grades 10-12.

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