$\displaystyle

\displaystyle f(x) = x^{4}-3x^{3}+3x^{2}

$

Find the open intervals for which concavity is open upwards.

$\displaystyle

\displaystyle f'(x)=4x^3-9x^2+6x

$

$\displaystyle

\displaystyle f''(x)=12x^2-18x-6

$

I set f''(x)=0

$\displaystyle

\displaystyle 6(2x^2-3x-1)=0

$

$\displaystyle

\displaystyle \frac{3\pm\sqrt{17}}{4}

$

So when I test the points to the left & to the right, and in-between, the positive intervals are:

$\displaystyle

\displaystyle (-\infty,\frac{3-\sqrt{17}}{4}],[\frac{3+\sqrt{17}}{4},\infty)

$

which is apparently wrong. Where did I mess up? Thanks!