Let the cubic polynomial f(x) = x^3 - px + q | vedclass

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

Vedclass · Accepted Answer

The cubic has a local minimum at $x = \sqrt{\frac{p}{3}}$ and a local maximum at $x = -\sqrt{\frac{p}{3}}$.

Vedclass · Answer

(A) Given the function $f(x) = x^3 - px + q$. To find the critical points,we find the first derivative: $f'(x) = 3x^2 - p$. Setting $f'(x) = 0$,we get $3x^2 = p$,which implies $x^2 = \frac{p}{3}$,so $x = \pm \sqrt{\frac{p}{3}}$. Now,we find the second derivative to determine the nature of these points: $f''(x) = 6x$. At $x = \sqrt{\frac{p}{3}}$,$f''(\sqrt{\frac{p}{3}}) = 6\sqrt{\frac{p}{3}} > 0$ (since $p > 0$). Thus,$f(x)$ has a local minimum at $x = \sqrt{\frac{p}{3}}$. At $x = -\sqrt{\frac{p}{3}}$,$f''(-\sqrt{\frac{p}{3}}) = -6\sqrt{\frac{p}{3}} Therefore,option $A$ is correct.

Let the cubic polynomial $f(x) = x^3 - px + q$ have three real roots,where $p > 0$ and $q > 0$. Which of the following is true?

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