So you add those two, you get minus 4, 12 minus 16 is minus 4, x to the sixth, all of that over x to the fourth plus 27 squared. And that is our second derivative. Now, we've done all of the derivatives, and this was actually a pretty hairy problem, and now we can solve for when the first and the second derivatives equal 0, and we'll have our candidate, well we'll know our critical points, and then we'll have our candidate inflection points and see if we can make any headway from there. So first let's see where our first derivative is equal to 0 and get our critical points, or at least maybe, also maybe where it's undefined. So this is equal to 0, if we want to set, the only place that this can equal to 0 is if this numerator is equal to 0. This denominator actually, if we are assuming we're dealing with real numbers, this term right here is always going to be greater than or equal to 0 for any value of x, because it's an even exponent, so this thing can never equal 0, right, because you're adding 27 to something that's non-negative, so this will never equal 0, so this will also never be undefined. So there's no undefined critical points here, but we can set the numerator equal to 0 pretty easily.