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University of North Texas



Problem 15 Medium Difficulty

If $ f $ is the function considered in Example 3, use a computer algebra system to calculate $ f' $ and then graph it to confirm that all the maximum and minimum values are as given in the example. Calculate $ f" $ and use it to estimate the intervals of concavity and inflection points.


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Video Transcript

in example, three. We craft the function X squared times X plus one cute all over X minus two squared Times X minus four. The fourth power just by doing a rough sketch of it and then using that rough sketch idea toe look at certain pieces of our ground so we can zoom and are zoom out to kind of look at particular overarching structure of the graph to find our minimum values. And we find that at about X's and the negative 20 X is equal to negative. 0.3 and 2.5. We have minimal mountains. So what we want to do it's used some kind of caste system to find the first and second derivatives. We want to use the graph of the first riveted to confirm that those minimum values we found our all the given, um, minimums and we didn't miss anything. And also we want to use the second derivatives graft to estimates for intervals of con cavity as well as any inflection points. Uh, so you might see from what f prime is what they say. An example. Three about trying to take the derivative of F is going to be way too time consuming. I mean, that is pretty big and bulky, so I can only imagine the elder, but we have to do to simplify it to that point. So just go ahead and plug and FX into your favorite derivative calculator, and you should get the same thing they that I have here on the board. Now, this first graph is just a overall graph of what it looks like for prime, but it doesn't really give us much information. So this second graf over here is when I went in and zoomed to the left of the graph and I made sure to go all the way out to at least negative 20. Get these other about use. So let's just start from left. So it does look like about negative 20. We will have a minimal because we're decreasing since the functions negative toe left of that point and then increasing after, which tells us we have a minimum there. Then So over at exit one level below Check mark underneath 20 Now around at negative one. Well, there it really doesn't do anything, does it just bounce it all and then our next point here. Well, that's at about negative 0.3. But you might notice that this isn't actually a minimum value, because we are increasing into the point and then decreasing after. So while negative 20 I'll go ahead is a minimum at X zero to negative 0.3. We actually have a maximum. And now over here at X is equal to zero. They didn't list that one in the problem. Well, let's just go ahead and see what that is. So x is equal to zero. Is that point right there? And so it is decreasing into the point and then increasing afterwards, which means this will be a minimum. And then over here, at about 2.5, which is our last value that we had, we can see that we are decreasing into the point, so decreasing into it and then increasing afterwards. And I know it's kind of hard to see, but at that point should kind of look like that for the subs were slightly to the left. And then you go slightly to the right. No, I'm sorry. Actually, uh, that point there should be like this because from zero we're going to a pause or after they were going to deposit value. And since we have no interception, anything it should be like. So it's a good idea to actually draw all those. So you can actually read him a little bit better. We got a nice enough grab so you don't make the same mistake. I wanted to do it first. So that actually says we're increasing into the point and then decreasing afterwards. Which would mean at this point, we have a max. So actually, two of these, even though we estimated them to the minimums, they're actually supposed to be local maximums of our function. Now that we've done that, we can go ahead and use thes second derivative two find our possible points of inflection and all of that at intervals of contact. Now, what I would have had it did was second derivative using our caste software. And this first craft is just a overarching graph of it. And the only information that we really will need to look at is to the left of, um are vertical ascent over X is equal to. So I went ahead and blew that information. And this is what I ended up getting for our interval, and I kind of went out kind of far for the xxx and it didn't really look like it had anything. So these I would say our all of our ex intercepts. So let's go ahead and find our intervals of con cavity. I'm gonna go ahead and get rid of this arrow just to get us a little bit of space so this function will be con cave up. So I'm gonna start from the left, so I'll start on this one here so we can start looking at everything. So when F double prime of X is strictly larger than here, so that's going to be from negative infinity to about negative four point 978 union with and then negative 1 to 0.543 union with and then negative 0.108 until our first intercept or ah, vertical accents and executed. And then the rest of our domain should also be greater than two. It would be too. 24 union for two. So then that tells us the rest of our interval should be where f double finals. Listen, zero. We're just going to be negative or 0.978 to negative one union live 0.5 for three two negative 0.10 and to remember, F double prime lesson zero means concave down. And when f double crime is strictly greater than zero, this is Kong Cave. Um