# 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?

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• Perhaps you should start by finding all the inflection points by finding the zeros of $f''(x)$ – PossiblyDakota Sep 29 '18 at 5:23

$$f(x) =\frac{a}{3}x^3 +(a+2)x^2+(a-1) x+2$$

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 - .