 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Concavity and Inflection Points - Problem 3

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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Find the second derivative of the function (do this by calculating the first derivative, then calculating the derivative of the first derivative). Then find the points where the second derivative is 0 or undefined.

To determine where the function is concave up or concave down, find where the second derivative is positive or negative, respectively. This can be done using the same techniques as we used to find the intervals of increase and decrease with first derivatives. The inflection points of a function are the points where the second derivative is 0 and there is a change from concave up to concave down.

Let's do another problem where we look at the concavity, and inflection points of a function. Here we're asked to determine intervals of concavity, and find inflection points for the function g(x) equals x times the quantity 5 minus x to the 2/3.

Now we've looked at this function before. In the previous problem we found that, it's derivative was 15 minus 5x over 3(5-x) to the 1/3. This will save us a little time, because we need to find the second derivative. So let's do that now.

We'll use the quotient rule, because this is in the form of a quotient. It's low d high. D high meaning the derivative of the top. So that's -5 minus high d low. That's going to be the derivative of this bottom here. It's 3 times, and the 1/3 comes out in front, times 5 minus x. I had to replace this exponent with one less. 1/3 minus 1 is -2/3. But because I don't have just an x inside, I need to use the chain rule. I have to multiply by the derivative of the inside part. So I multiply by -1. This completes the derivative of the low part. Then I have to have over this square of what's below. So this is going to be 9 times 5 minus x to the 2/3.

Now let's simplify the numerator just a little bit. So I have 3 times -5, -15 times 5 minus x to the 1/3. I have a (5-x) to the -2/3. I have 3 times a 1/3 is 1 times -1 is -1, positive (15-5x) times (5-x) to the -2/3.

Now technically, this is a complex fraction, which means it has fractions inside of fractions. Because this negative exponent indicates that this is a fraction. Now I can get rid of that by multiplying the top, and bottom by (5-x) to the +2/3, like this. So (5-x) to the 2/3 on top, (5-x) to the 2/3 on bottom. That will get rid of my negative exponents. That will actually simplify this a lot.

First, let's take a look at what we're going to have in the denominator. When we multiply the denominators through we'll have (5-x) to the 2/3 times (5-x) to the 2/3. That's (5-x) to the 4/3. We'll still have this 9 here. In the numerator, we still have -15, we have (5-x) to the 1/3 times (5-x) to the 2/3. Add the exponents, (5-x) to the 1. Plus, we have this (15-5x), and then we have (5-x) to the -2/3 times (5-x) to the +2/3, becomes (5-x) to the 0 which is 1.

Now this will simplify a lot in the numerator. Let's see what we have. We've got -75 plus 15, that's -60. We have 15x minus 5x, that's +10x. Very nice. In the denominator we have 9 times (5-x) to the 4/3. That's our second derivative.

Let's take this up here, because the next step is to find where the second derivative is 0. So g''(x) is -60 plus 10x over 9 times (5-x) to 4/3. We're looking for where the second derivative is 0, or where its undefined. Now here it is possible for the second derivative to be undefined when x is 5. When x is 5, we get 0 in the denominator, and the second derivative will be undefined. So x equals 5 is a critical point, or a point where the second derivative is 0, or undefined.

We can also get the numerator to equal 0 when x is 6. So x equals 6 will make the second derivative equal 0. So these are both points we need to consider, because these are both points where the second derivative could change sign. Now we're going to make sign chart. We should put both of these numbers 5, and 6 on our sign chart.

Let's remember that at x equals 6, the second derivative is 0. At x equals 5, the second derivative is undefined. I'll put a 'u' for that. Still these points divide the number line into three intervals. I need to check what's happening to g'' in each of these intervals. So let me take a look at a point to the left of 5 like 4.

When I plug 4 in, I get 40 here minus 60. See I get a negative number, -20 on top. When I plug 4 in here, I get 5 minus 4 which is 1 to the 4/3. It's going to be positive. So negative over positive, negative.

When I plug in something between 5, and 6 like 5.5, I get 55 minus 60, still that's a negative on top, 5 minus 5.5 is negative. But when you raise it to the 4/3, it becomes positive. Remember raising the 4/3, is like first raising to fourth power which will make a negative positive, then taking the cube root. So it will be positive. So again, 55 minus 60 is negative, but the denominator is positive.

Now let's take a look at 6. When we go to the right 7, we get 70 minus 60, that's positive. 5 minus 7, negative, but raised to the fourth root will become positive. Positive over positive is positive. What do we have? We have that g is concave down both here, and here, and concave up here. So let's write that down. It's concave down on the interval from negative infinity to 5, that's to the left of 5. It's also concave down between 5 and 6. It's concave up from 6 on, because the second derivative is positive.

Now as for inflection points, 5 doesn't turn out to be an inflection point, because there is no change in concavity. It goes from concave down, to concave down, not in an inflection point.

So again it's very important that you don't just assume that these end up being inflection points. You really have to check on your sign chart. This is not an inflection point. But x equals 6 is an inflection point, because the second derivative goes from negative to positive indicating that we're going from concave down to concave up. So we have an inflection point at x equals 6. Two intervals where the function's concave down, and one where it's concave up.