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# Curve Sketching with Derivatives - Problem 4

###### 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.

In order to sketch a graph, you need to know a function's critical points and inflection points. To find the critical points of the function, calculate the first derivative and find its zeros. Then, find the value of the function at those points (by plugging them in to your original function). These points will be your relative maxima and/or minima. Then, to find the inflection points, find the zeros of the second derivative. Remember that these are points at which the function may change from concave up to concave down, and vice versa--this can help you visualize the graph better.

I want to do a really challenging curve sketching example. I’m asked to graph the function f(x) equals 144 over x² plus 12. The extra simplification here is that this graph has an asymptote. So I’m going to have to find that in the process of my planning for the graph. I’m going to go through each of the steps you see here in detail.

In previous problems, I have found the derivative and I did that ahead of time. This time I’m actually going to go through the process of finding f' and f'', making the sign chart, finding the asymptote or asymptotes and plotting special points. You’ll see every step in this problem.

First step in the graphing process, is to take the first and second derivatives. I’m going to do that right now. This function’s going to require the quotient rule. F' equals, remember my memory device it’s low d high. And the derivative of the high part is going to be zero here, minus high d low, and that’s the derivative of x² plus 12x, it’s 2x. Over the square of what’s below. That’s going to give me -288x on top and on bottom x² plus 12².

Now you’ll have to take the derivative of this to get the second derivative and I also need to use the quotient rule. So f''(x). Let me just write f'(x) up here; again that was -288x over x² plus 12². Let me move this down just a little bit. F''(x), quotient rule.

So it’s low d high x² plus 12² times the derivative of the high part, that’s -288 minus high d low. That’s going to be -288x times the derivative of this. That’s going to require the chain rule. It will be 2 times (x² plus 12)² to the 1 times the derivative of x² plus 12. 2(x² plus 12) to the 1 and the derivative of x² plus 12 is 2x. All that over the square of what’s below. So over x² plus 12 to the 4th.

You can see that one thing that will make this simpler right away, is if I cancel the common factor of x² plus 12, with one of these in the denominator. Both of these terms, this long term here and this one here,, have a factor of x² plus 12. I’m going to get rid of that. Get rid of this one and one of these with one of these.

Let’s look at what remains. We’ve got -288x² plus 12 times -288. And then we have 2 times 2, 4x, minus, minus, plus 4x² times 288. That’s it, all over (x² plus 12)³. Let’s take a look at this. I’ve got -288 x² plus 4x² times 288. I think I wanted to keep that 288 outside. I’m going to leave it outside. I have -x² plus 4x², that’s 3x². And then here if I keep the 288 out, I have a -12. (x² plus 12)³.

Now there’s one more thing I could do to make this a little bit simpler for me to deal with in the future parts of this problem. I want to pull a factor of 3 out and combine it with a 288. I’ll get 864x² minus, I pull a 3 out, I get a 4 here. Over (x² plus 12)³. Now I have the first derivative and the second derivative and I’m ready to make a sign chart.

We’re ready to make a sign chart for our f' and f''. First thing we have to do is find critical points. Let’s take a look at our derivatives. One of the things that makes these derivatives look complicated is the crazy denominators we have to deal with. Remember x² plus 12 is a sum of squares. It’s always going to be positive. You can’t plug a number in to make it negative. And so there is not going to be any sign changes associated with the denominator. We simply don’t have to worry about it. This will always be positive. The only place that this derivative can ever equal zero, is at x equals zero. One critical point is x equals zero. That’s the only critical point for the first derivative.

Let’s take a look at the second derivative. Again x² plus 12 always positive, so this thing is never undefined, the only critical points are hiding in this factor, x² - 4. You can see that x equals 2 or -2 are going to make this zero. so those are two more points we need to consider; x equals 2, x equals -2.

Let’s make our sign chart for this numbers. I’ll put in -2, 0 and 2. I recopied my derivatives up here so I can do my calculations. Let’s remember that at x equals zero the first derivative is zero. And then we need to figure out what happens to the left, and what happens to the right.

On the left side, I could plug in -2 and I don’t even really need to know what -288 times -2 is, I just need to know it’s positive. Negative times a negative so it’s positive. Divided by a positive, that will be positive. And then on the right when I plug in 2, I have a negative times 2 times a positive, that’s negative over a positive, this will be negative. It’s all I need to know.

Now down here, the second derivative has zeros here and here at negative 2 and 2. What happens to the left? Let’s try something like -3. When I plug in -3, I’m going to get 9 minus 4, that’s positive, times positive, positive on top. And this is always positive, so it’s going to be positive over here.

Actually let me observe that 3 is going to be the same situation. I plug in 3; 9 minus 4, positive, times positive over positive, that’s positive. A test between -2 and 2, I’ll plug in zero. When I plug in zero I get a negative number times a positive over a positive, this will be negative in between.

So really quickly I can use this sign chart to give me some idea of what the shape of the graph is going to be. Here, where the first and second derivatives are positive, the graph’s going to be increasing and concave up. So something like this.

Here where the first derivative’s positive, the second I negative, the graph’s increasing and concave down. So something like this. Here the first and second derivatives are both negatives, decreasing concave down. And here first derivative’s negative, second derivative’s positive so the graph’s decreasing and concave up. Something like that. Let’s keep this in mind when we’re graphing in the next part. We also have to come up with key points and asymptotes.

So now we’re ready to graph key points, find asymptotes and graph this function. Let’s take a look at our sign chart. And remember that, it looks like an x equals zero, we’re going from increasing to decreasing so we’re going to have a local maximum. You can see that here. Let’s plot that. Let’s find its coordinates. It’s going to be f(0), really easy to calculate if you go up here, 144 over 0² plus 12. So 144 over 12 which is 12. We’re looking at f(0) equals 12. That’s our local max.

Let’s take a look at this point which looks like it’s going to be an inflection point. It’s running from concave down to concave up. Plug in 2 to this function I get 144, over 2² 4 plus 12. 144 over 16. That’s f(2), 144 /16, that’s 9. Let’s observe that when we plug in 2, which has a similar thing happening, changing concavity, -2 should give me the same answer. So f(-2) also 9.

These are my two inflection points, there is my local max. My local max also happens to be an intercept, so, doubly important. Now I just need to talk about the asymptote. Let’s take a look at our function here. Recall, to find the horizontal asymptotes, you need to consider the limit as x approaches infinity and the limit as x approaches negative infinity.

So limit as x approaches infinity. We’re just going to think this through a little bit here. 144 over x² plus 12. Think about what happens to the denominator as x goes to infinity. X² is going to go to infinity as well, and x² plus 12, even more so. So the denominator’s getting huge as x goes to infinity. Think about what happens to a fraction when its denominator gets bigger and bigger. The fraction gets smaller and smaller, closer to zero. So this limit is going to be zero.

And, that’s going to be true if I let x go to negative infinity as well. Because this x², it’s going to respond the same to x negative infinity as x going to positive infinity. So both limits are going to be zero and we’re going to have our graph approaching zero in both directions. Let’s think about that as we’re graphing.

Now I’ll throw in a y axis here. Remember (0, 12) was our local max, right about here. I’ll put -2 and 2 in. Remember that the value is 9 at those points. So about ¾, here and here. In between -2 and 2, remember it was concave down. These are inflection points. It’s concave down here and here. And it’s going to have a horizontal tangent here. So something like this, and then over here it’s concave up, but decreasing. Remember the asymptote. It’s going to approach the x axis. Same thing in this direction. Approach the x axis.

That’s it. I’ll just apply some points. Teachers love that on the test, you got to plug a lot of points, show what you know about the function. We’ve got two inflection points, our local max and our asymptotic behavior. There’s our graph of f(x) equals 144 over x² plus 12.

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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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