Concavity and Inflection Points - Problem 1

I want to talk about concavity and inflection points. I have an example here. Determine the intervals of concavity, and find inflection points for the function g(x) equals 3x to the fourth minus 20x³ plus 17.

Now the first step, is to find the first and second derivatives. I'm really interested in the second derivative. I have to find the first to get the second. First derivative is going to be 12x³ minus 60x², and the derivative of this term is 0. That's my first derivative.

The second derivative. Well, the derivative of the 12x³ is 36x², and the derivative of 60x² is 120x. Now whenever you're dealing with derivatives you want to find inflection points, concavity. You want to factor the derivatives as much as possible. Here I have a common factor in both of these terms of 12x. So I'm left with 3x here minus 10. So I have a nicely factored second derivative.

This takes me to the next step. Find where g''(x) is 0. Well, g''(x) was 12x times 3x minus 10. Now remember this is like when you're looking for critical points. You do want to find where the second derivative is 0, but you also want to take into consideration where its undefined if it is undefined. This second derivative is not undefined anywhere, so we don't have to worry about this example.

Now where is this 0? Well, since it's factored, it's really easy to tell. 0 when x equals 0, or when x equals 10/3. Now you don't want to jump to conclusions too soon, and say these are the inflection points. They may end up being the inflection points, but you actually need to see that change in concavity to be sure that they're inflection points. We'll get to that in a bit.

Now we make a sign chart for g''. The real purpose of finding these points was so we can mark them on our sign chart. These are points where the sign of g'' may change, but it doesn't have to. 0, and 10/3. Now what we know about these points is that g'' is 0 at each of those. So we want to test points to the left, in between, and to the right to see whether g'' is positive or negative.

So let's try -1. I'll plug -1 in here, and I get a negative number, -12. Then I get 3 times -1 minus 10, that's going to be negative as well. So negative times negative, positive. It doesn't even matter what the value is. I just need to know it's positive.

Between 0, and 10/3, given my 10/3 is like 3, and a 1/3. So let's try 1. 12 times 1 is 12, that's positive. Then I get 3 minus 10, that's negative, so positive times negative, negative. Then to the right of 10/3, I can try 12/3, which is 4. So I'll plug in 4, and I get 48. Then 12 minus 10, that's positive, that's positive. So it's a product of positive numbers which is positive.

Now what I need to analyze here is whether the function is concave up, or con cave down. Here, because second derivative is positive, g is going to be concave up. It's also positive here, so it will be concave up from negative infinity to 0, and from 10/3 to infinity. So that's where it's concave up. It's concave down where the second derivative is negative, and that's between 0, and 10/3. So intervals where it's concave up from negative infinity to 0, and 10/3 to infinity. It's concave down from 0 to 10/3.

What about inflection points? Well, remember you needed to see that change in concavity. We do have a change in concavity here. It goes from concave up, to concave down at x equals 0. So that's inflection point. It goes from concave down to concave up at 10/3. So another inflection point.

So we do in fact have inflection points at x equals 0, and x equals 10/3. You really need to check that there actually is a change in sign, in order for a point to be an inflection point. Don't just assume that it is, just because the second derivative is 0. It's good to look at the sign chart, and look for that change in sign.