Let $f(x)=x^3-px+q,$ where $p>0$ and $q>0$.

We need to determine where the cubic has local maxima and minima.

1. Find the critical points

Differentiate: $f'(x)=3x^2-p.$

Set $f'(x)=0$: $3x^2-p=0$ $x^2=\frac p3$ $x=\pm \sqrt{\frac p3}.$

So the stationary points are at $x=\sqrt{\frac p3} \quad \text{and} \quad x=-\sqrt{\frac p3}.$

2. Use the second derivative test

Differentiate again: $f''(x)=6x.$

Now test each critical point.

At $x=\sqrt{\frac p3}$

$f''\left(\sqrt{\frac p3}\right)=6\sqrt{\frac p3}>0.$ Since $f''>0$, the function has a local minimum here.

At $x=-\sqrt{\frac p3}$

$f''\left(-\sqrt{\frac p3}\right)=-6\sqrt{\frac p3}<0.$ Since $f''<0$, the function has a local maximum here.

3. Match with the options

• A: minima at $\sqrt{\frac p3}$ and maxima at $-\sqrt{\frac p3}$ ✅
• B: reversed ❌
• C: minima at both points ❌
• D: maxima at both points ❌

Hence the correct option is: $\boxed{A}$

4. About the condition "three distinct real roots"

The given condition is not actually needed to determine the nature of the stationary points. The maxima/minima locations depend only on the derivatives of $f(x)=x^3-px+q$, and since $p>0$, the critical points are always real and their nature is fixed as above.