Point of negative inflection 
Find the range of values of a for which the function $f(x) = ax^3/3 +(a+2)x^2+(a-1)x + 2$ possesses a negative point of inflection. 

My attempt
I differentiated it twice and equated it less than $0$ to get $x < \frac{-(a+2)}{a}$. This isn't giving me any range of $a$. Any hint?
t

$f(x) =\frac{a}{3}x^3 +(a+2)x^2+(a-1) x+2$
I think you are supposed to find the intervals for which the function is concave-down? Note that the first derivative of the given function will allow you to find local mins and maxes:
$\frac{df(x)}{dx} = ax^2 + 2(a + 2)x + (a-1) = 0$
$x = \frac{-(a+2) +- \sqrt{4a + 5}}{a}$
Check the concavity (sign of any point in $d^2 f/dx^2$) in the regions
$(-inf.,\frac{-(a+2) - \sqrt{4a + 5}}{a})$,
$(\frac{-(a+2) - \sqrt{4a + 5}}{a},\frac{-(a+2) + \sqrt{4a + 5}}{a})$,
$(\frac{-(a+2) + \sqrt{4a + 5}}{a}, inf)$
Second derivative is (as you found)
$\frac{df^2(x)}{dx^2} = 2ax + 2(a + 2)$
Check the sign of that function when evaluated at any convenient point in those intervals. You want - .
