Using the first and second order derivatives to find the critical points of this function has left me very confused -- they don't mesh at all with the critical points on the graph of the function (as displayed by a graphing calculator).

$\displaystyle f(x) = 3x^{4} - 2x^{3} - 12u^{2} + 18x -5$

$\displaystyle f^{\prime}(x) = 12x^{3} - 6x^{2} - 24x + 18$

$\displaystyle f^{\prime}(x) = 6(2x-2)(x-1)(x+1.5)$

$\displaystyle f^{\prime \prime}(x) = 36x^{2} - 12x -24$

$\displaystyle f^{\prime \prime}(x) = 12(3x + 2)(x -1)$

So, that give potential critical points of x= 1, and -3/2 for relative extremas and x=-2/3 for a point of inflection.

The calculator displays critical points (min, max, min, respectively) of x= 3, 0.6861393, and -2.186138.

Why the discrepancy?