Suppose the cubic x^3 - px + q has three distinct real roots where p0 and q0.

Question

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?

Accepted Answer

Let $y = {x^3} - px + q$

$$ \Rightarrow {{dy} \over {dx}} = 3{x^2} - p$$

For $${{dy} \over {dx}} = 0 \Rightarrow 3{x^2} - p = 0$$

$$ \Rightarrow x =  \pm \sqrt {{p \over 3}} $$

$${{{d^2}y} \over {d{x^2}}} = 6x$$

$${\left. {{{{d^2}y} \over {d{x^2}}}} \right|_{x = \sqrt {{p \over 3}} }} =  + ve\,\,\,\,$$ and

$$\,\,\,\,\,\,\,\,\,\,$$ $${\left. {\,\,\,{{{d^2}y} \over {d{x^2}}}} \right|_{x =  - \sqrt {{p \over 3}} }} =  - ve$$

$$\therefore$$ $y$ has ninima at $$x = \sqrt {{p \over 3}} $$

and maxima at $$x =  - \sqrt {{p \over 3}} $$

Solutions · 1