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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

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  • $\begingroup$ Perhaps you should start by finding all the inflection points by finding the zeros of $f''(x)$ $\endgroup$ – PossiblyDakota Sep 29 '18 at 5:23
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$f(x) =\frac{a}{3}x^3 +(a+2)x^2+(a-1) x+2$

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 - .

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