# Points of Inflection

• October 10th 2012, 11:19 PM
arcticreaver
Points of Inflection
the question is find the points of inflection of f(x) = 1/2x^4+x^3+6x^2-2x-4

now my understanding is that to find points of inflection, i would need to find the 2nd derivative first, right?

this is what i have for the 2nd D - 6x^2+6x+12. then you would factor so 6(x^2+x+2), i can't really go on after this because you can't really factor it out, right? so there is points of inflection for this equation? i am guessing there is not...?
• October 10th 2012, 11:40 PM
MarkFL
Re: Points of Inflection
You are right, but for the wrong reason. Just because you cannot factor the quadratic doesn't mean it has no real roots. In this case there are no real roots, but you should probably use the quadratic formula to demonstrate this.
• October 10th 2012, 11:42 PM
arcticreaver
Re: Points of Inflection
so should i work out the quadratic equation and say something like "there is no points of inflection"? or that points of inflection does not exist for this equation?
• October 10th 2012, 11:45 PM
arcticreaver
Re: Points of Inflection
actually, then only x=0 right? so would the inflection point be just (0,0)?
• October 10th 2012, 11:52 PM
MarkFL
Re: Points of Inflection
No, since the 2nd derivative has no real roots, then there are no places where it can change its sign, which is what is required for a point of inflection. The 2nd derivative must go from negative to postive or positive to negative at a point of inflection, but this function as no such points.

We need only analyze the discriminant of:

$x^2+x+2$

which is:

$1^2-4(1)(2)=-7$

Since the discriminant is negative, the quadratic has complex roots, so it has no real roots, hence there can be no points of inflection.