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# Concavity and Inflection Points - Problem 2

###### 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.

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.

We're studying concavity and inflection points. Here is a problem. Determine intervals of concavity and find inflection points where the function h(x) equals 9x times e to the -x over 3. So for this problem, we're going to need to use the product rule to find the derivatives.

First, h'. Now the product rule says first times the derivative of the second. So that first is going to be 9x, and the derivative of the second is going to be e to the -x over 3 times -1/3. Don't forget the chain rule on that. So it's going to be times -1/3e to the -x over 3. So that's the derivative of e to the -x over 3 plus the second e to the -x over 3 times the derivative of the first which is 9. So times 9.

We should probably simplify, and factor this completely before we try to take the second derivative. Otherwise we'll do more work than we have to. This -1/3 times 9x is going to give -3x. -3x e to the -x over 3 plus 9 times e to the -x over 3. Now we see that this is going to factor very nicely. (-3x plus 9) e to the -x over 3. That's the first derivative.

So the second derivative will also involve the product rule, because this is in the form of a product. Once again, it's first times derivative of the second, plus the second times the derivative of the first. So it's -3x plus 9 times, and the derivative once again of e to the -x over 3 is -1/3e to the -x over 3, plus the second times the derivative of the first. That's e to the -x over 3 times. The derivative of the first is just -3.

Now long story short, this is going to factor very much like the first derivative did. There's going to be an e to the -x over 3 in both of these terms. Let me just write that e to the -x over 3 times, and then a bunch of stuff. Here I'm going to have -3x times -1/3. That's just x. I have 9 times -1/3. That's -3. Over here I also have a -3, so minus another 3. That's going to be x minus 6 times e to the -x over 3. That's my second derivative.

So I'll take this second derivative, and I'll pull it up here, because my next step is to find out where the second derivative is 0. x minus 6 times e to the -x over 3. Find out where the second derivative is 0, or undefined. Now this will never be undefined. There is no reason why this will be undefined. There is no division by 0. No danger of that. No danger of negatives, and square root.

So we have to look for where this thing equals 0. Now e to the -x over 3 will never equal 0. This will always be a positive number. So we don't need to worry about this. Only this can never equal 0. This equals 0 at x equals 6. So that's going to be one of the points that we're going to have to look at, when we're making our sign chart. That's step three.

Make a sign chart for h''. This is h''. I've got to put 6 on my sign chart, that's where h'' is 0. Then I have to test to the left and to the right. So testing to the left, I'll pick something like why don't I just pick 0? I'll plug 0 in. I'll get a negative number here, e to the 0, e to anything is positive. So negative times positive is negative. Then I'll pick something to the right like 7. 7 minus 6 is positive, it's 1. E to the -7/3, again e to anything is positive. So we have a positive times positive, this is positive.

Now what this tells me is about the concavity of h. H is going to be concave down to the left of 6, concave up to the right. So I can fill that in right now. H is concave up on the interval from 6 to infinity. It' s concave down on the interval from negative infinity up to 6.

Now what about inflection points? Well, there is a sign change here. So x equals 6 is an inflection point. It's the only one. X equals 6 is the inflection point. So once again, where the second derivative is negative, your function h, is going to be concave down. Where the second derivative is positive, h will be concave up. And because there's a sign change here, the graph of h will have an inflection point at x equals 6.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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