Suppose the cubic x^3 - px + q has three dist | vedclass

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(D) Let $f(x) = x^3 - px + q$.To find the critical points,we calculate the first derivative: $f'(x) = 3x^2 - p$.Setting $f'(x) = 0$,we get $3x^2 = p$,

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The cubic has minima at $\sqrt{\frac{p}{3}}$ and maxima at $-\sqrt{\frac{p}{3}}$.

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(D) Let $f(x) = x^3 - px + q$.To find the critical points,we calculate the first derivative: $f'(x) = 3x^2 - p$.Setting $f'(x) = 0$,we get $3x^2 = p$,which implies $x = \pm \sqrt{\frac{p}{3}}$.Now,we find the second derivative: $f''(x) = 6x$.Evaluating at the critical points:For $x = -\sqrt{\frac{p}{3}}$,$f''(-\sqrt{\frac{p}{3}}) = -6\sqrt{\frac{p}{3}}  0$). Thus,there is a local maximum at $x = -\sqrt{\frac{p}{3}}$.For $x = \sqrt{\frac{p}{3}}$,$f''(\sqrt{\frac{p}{3}}) = 6\sqrt{\frac{p}{3}} > 0$ (since $p > 0$). Thus,there is a local minimum at $x = \sqrt{\frac{p}{3}}$.Therefore,the cubic has a maximum at $-\sqrt{\frac{p}{3}}$ and a minimum at $\sqrt{\frac{p}{3}}$.

Suppose the cubic $x^3 - px + q$ has three distinct real roots where $p > 0$ and $q > 0$. Then which one of the following holds?

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