 ###### 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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# Curve Sketching with Derivatives - 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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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.

Let’s do another curve sketching example. Here I want to graph h(x) equals 9x times e to the –x over 3. I have the first and second derivatives already taken and factored here. That’s the first step ,in any curve sketching problem. Make sure you take the first and second derivatives, and factor them. I want to get sign chats for these derivatives.

Let me just take a look at the derivatives and see what the critical points are. I’m observing, when I look at the first derivative that, x equals 3 is going to make my derivative equals 0. So that’s a critical point. And that’s it. This expression e to the –x over 3 that can never equal 0, and this thing is never undefined. So that’s the only critical point. H'' is going to equal 0 when x is 6. So that’s another point I need to consider. So my two critical points, 3 and 6.

Let me make my sign chart. I’ll have h'' here, h' here. And remember at x equals 3, h' is 0. Now let’s see what happens to the left and to the right. I’ll test a point like x equals zero. I’ll plug in zero here and I’ll get 9 times, and this always positive, so it’s a positive times a positive. It’s going to be positive over here.

And then to the right of 3, I could try anything; 4. -3 times 4 is -12 plus 9 that’s -3. This is again always positive so it’s a negative times a positive, negative. The second derivative is zero at 6. That’s what we found before. Let’s test it to the left. Let’s plug 3 in and see what happens with the second derivative. You can see immediately this is going to be negative again this is always positive. So we have a negative second derivative. In fact I want to just put negatives all the way across. It’s negative all the way up until x equals 6.

What happens afterwards? Let’s plug in 7. 7 minus 6 is positive. E to the -7 is always positive. So this is going to be positive and let’s remember that the derivative is negative to the right of 3. It’s negative forever to the right of 3. So just continue that so you know what’s happening on this interval.

Let me draw little curves to remind myself what the shape’s going to be like. What does a curve that’s increasing and concave down look like? Well something like this. And then decreasing concave down, something like this. Now here it’s decreasing and the second derivative is 0. So it straightens out for a moment and then the concavity becomes concave up. So something like this. When you piece these together you have a nice little inflection point. That point, the inflection point and what looks like a local maximum, these are key points that I’ll want to graph.

So h(3), that’s one key point, let me take a look at my function. 9 times 3 is 27, e to the -3/3, that’s 27e to the -1. And I calculated this before, it’s approximately 9.9. I have to calculate h(6), going back to my function, 9 times 6, 54. E to the -6 over 3, e to the -2. So this is 54e to the -2 which is approximately 7.3. One other point I should plot just because it’s easy, x equals zero is going to be an intercept. When you plug in zero, you will get zero. So h(0) equals zero. That’s a point I should plot.

I mark my axis. I have x equals 0, x equals 3, and x equals 6 to consider. At x equals 0, the graph’s going to pass right through the origin and it’s going to reach a maximum of about 9.9 and 3. So this will be the point (3, 9.9) approximately. As we remember, it increases concave down up to this point, so the left part of the graph will look like this. It’s always increasing and concave down to the left of 3.

And then, between 3 and 6, it’s decreasing and concave down. It does that until it gets down to the value at h(6) of about 7.3. So it doesn’t go down very much. So from here down, just a little bit to (6, 7.3). And then it’s decreasing and concave up.

And let’s make the observation that, just going back to the function for a moment, this function, once x is positive, this will always be positive. No matter how large x gets, this will be positive. Now it’s decreasing after x equals 6 and concave up. It’s going to approach the x axis asymptotically, or it will stop short of the x axis. But you can actually prove, that this thing decreases very fast and overpowers this 9x, and causes the values of the function to go to zero. So decreasing and concave up. Something like that. I’ll throw in a y axis right there and I’m done. I got a pretty good graph of h(x) equals 9x times e to the –x over 3.