From a past exam. For the function y=x(x-3)^2/5, is there trick to find the inflection point? Doing the second derivative takes a while and gives messy results, and the space they give me to work out the problem is small and can't fit everything in there comfortably, which makes me think that there's a shorter way, but if there is its over my head right now.
2026-04-19 03:18:40.1776568720
is there an easy way to find the inflection point of this function?
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The first derivative is $$y'=\dfrac15\bigl((x-3)^2+2x(x-3)\bigr)=\frac35(x-3)(x-1) $$ Now the inflection point correspond to the minimum of this quadratic polynomial, and the extremum of a quadratic polynomial is attained at the midsum of its roots, i.e. at $\color{red}{2}$ in the present case.