Key Concepts and Formulas

• First Derivative Test: Critical points are found where $f'(x) = 0$ or $f'(x)$ is undefined. These points are potential locations for local maxima or minima.
• Second Derivative Test: If $f'(c) = 0$ and $f''(c) > 0$, then $f(x)$ has a local minimum at $x = c$. If $f'(c) = 0$ and $f''(c) < 0$, then $f(x)$ has a local maximum at $x = c$. If $f''(c) = 0$, the test is inconclusive.

Step-by-Step Solution

Let the given cubic function be $f(x) = x^3 - px + q$.

Step 1: Find the First Derivative We find the first derivative of the function $f(x)$ with respect to $x$. $f'(x) = \frac{d}{dx}(x^3 - px + q)$ $f'(x) = 3x^2 - p$

• Why this step? The first derivative $f'(x)$ represents the slope of the tangent to the curve $f(x)$ at any point $x$. At local maxima or minima, the tangent line is horizontal, meaning its slope is zero. Finding the first derivative allows us to find the critical points.

Step 2: Find the Critical Points To find the potential locations of local extrema, we set the first derivative equal to zero: $f'(x) = 0$ $3x^2 - p = 0$ Solving for $x$: $3x^2 = p$ $x^2 = \frac{p}{3}$ $x = \pm \sqrt{\frac{p}{3}}$ These are our critical points.

• Why this step? Setting $f'(x)=0$ helps us identify the points where the function's rate of change is momentarily zero. These critical points are candidates for local maxima or minima. The problem states $p>0$, ensuring that $\frac{p}{3}$ is positive and $\sqrt{\frac{p}{3}}$ is a real number. This guarantees the existence of two distinct real critical points.

Step 3: Find the Second Derivative Next, we find the second derivative of the function by differentiating $f'(x)$: $f''(x) = \frac{d}{dx}(3x^2 - p)$ $f''(x) = 6x$

• Why this step? The second derivative helps us determine the concavity of the function at the critical points. This concavity is what distinguishes a local maximum from a local minimum.

Step 4: Apply the Second Derivative Test at each Critical Point

Now, we evaluate the second derivative $f''(x)$ at each of our critical points:

• For the critical point $x = \sqrt{\frac{p}{3}}$: Substitute $x = \sqrt{\frac{p}{3}}$ into $f''(x) = 6x$: $f''\left(\sqrt{\frac{p}{3}}\right) = 6\left(\sqrt{\frac{p}{3}}\right)$ Since $p > 0$, we know that $\sqrt{\frac{p}{3}}$ is a positive real number. Therefore, $6\left(\sqrt{\frac{p}{3}}\right)$ will be positive. $f''\left(\sqrt{\frac{p}{3}}\right) > 0$

  • Conclusion: According to the Second Derivative Test, since $f''\left(\sqrt{\frac{p}{3}}\right) > 0$, the function $f(x)$ has a local minimum at $x = \sqrt{\frac{p}{3}}$.

• For the critical point $x = -\sqrt{\frac{p}{3}}$: Substitute $x = -\sqrt{\frac{p}{3}}$ into $f''(x) = 6x$: $f''\left(-\sqrt{\frac{p}{3}}\right) = 6\left(-\sqrt{\frac{p}{3}}\right)$ Since $p > 0$, we know that $-\sqrt{\frac{p}{3}}$ is a negative real number. Therefore, $6\left(-\sqrt{\frac{p}{3}}\right)$ will be negative. $f''\left(-\sqrt{\frac{p}{3}}\right) < 0$

  • Conclusion: According to the Second Derivative Test, since $f''\left(-\sqrt{\frac{p}{3}}\right) < 0$, the function $f(x)$ has a local maximum at $x = -\sqrt{\frac{p}{3}}$.

Common Mistakes & Tips

• Remember the conditions: The condition $p > 0$ is crucial to ensure the critical points are real.
• Sign of the second derivative: A positive second derivative indicates a local minimum, while a negative second derivative indicates a local maximum.
• Second Derivative Test Limitations: If $f''(c) = 0$, the Second Derivative Test is inconclusive, and the First Derivative Test should be used.

Summary

By applying the Second Derivative Test, we found that the cubic function $f(x) = x^3 - px + q$ has a local minimum at $x = \sqrt{\frac{p}{3}}$ and a local maximum at $x = -\sqrt{\frac{p}{3}}$. This corresponds to option (A).

The final answer is $\boxed{A}$.