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Let X be a continuous r.v., with pdf $f_X(x) = kx(1-x), 0 < x <1$

I evaluated k = 6 and found the cdf $F_x(x) = 3x^2 - 2x^3$, but then I am asked to find the median. The equation $F_x(x) = 1/2$ brings me to $-4x^3 + 6x^2 -1 = 0$, for which I don't know any quick way to find its root (but I randomly plugged in $x = 1/2$ and got my answer). The question is, apart from the tedious cubic formula, is there another quick trick to solve general cubic equations, especially when I am not looking for integer solutions?

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Cubic equations are indeed tedious.

I guess you are expected here to realize that the pdf is symmetric around $x=1/2$ : that is, $f_X(x-\frac12)=f_X(x+\frac12)$. Hence, the median is $1/2$, because $\int_0^{1/2} f_X(x) dx =\int_{1/2}^1 f_X(x) dx$

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  • $\begingroup$ I see now, but I didn't know that the graph was symmetric in advance, and I suspect that examining the function (find roots, first and second derivatives and their roots, potential asymptotes etc) is too farfetched for the purposes of this question. $\endgroup$ – JBuck Apr 19 at 21:17
  • $\begingroup$ Which brings this question to my mind: Is every 2nd degree polynomial symmetric around its extremum? $\endgroup$ – JBuck Apr 19 at 21:19

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