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University of North Texas



Problem 9 Medium Difficulty

Produce graphs of $ f $ that reveal all the important aspects of the curve. Estimate the intervals of increase and decrease and intervals of concavity, and use calculus to find these intervals exactly.

$ f(x) = 1 + \dfrac{1}{x} + \dfrac{8}{x^2} + \dfrac{1}{x^3} $


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Video Transcript

you want to produce graphs of the function F of X is equal to one plus one over X plus eight rex word plus one over excuse that reveal all the important aspects of that curve. So we also want to estimate the intervals where the functions increasing, decreasing as well as any intervals of con cavity. And then we want to use calculus to find these intervals. Exactly. All right, so I first just went ahead and graft, whatever Becks look like. Oh, when I just plug it into my calculator right away, and so that doesn't really tell us much. So let's see if we can go ahead and use the 1st 2nd derivatives here to maybe get a better viewing window so you could just go ahead and zoom all the way out if you wanted to estimate these. But I'm just gonna go ahead and do it this way. Toe, get a nice viewing window. So, um, I already have the person taking derivatives here upon it, and I also found what I believe to be the ex intercepts. And we can use these values along with the fact remember, that at X is equal to zero. This function is undefined. So we have a horizontal ass in two vertical ass. I'm totally should say. Actually, at X is equal to zero for all three of these functions, since it's undefined at X is equal to zero. All right, so let's go ahead and find the intervals of increased decrease and comm cavity. So over here, so we know when F prime of X is strictly larger than zero. This will tell us where the function is increasing. So to the left of negative 0.19 it looks like F is going to be positive. So from negative infinity to negative 0.19 we would go ahead and say it's increasing and then it doesn't look like it's increasing anywhere else. And so then that would tell us F Primal Dex would be strictly less than zero on the remainder of our interval here. So from negative 0.19 to 0 union zero to and go to my fellow, then the next thing we want to do is to go ahead and look at the second derivative, so double prime of X is strictly larger than zero. Tells us where the function will be conking out. And just like before we have this X intercept or the vertical asking Joe at X zero. And so this is gonna be from negative infinity to negative 0.2 by three, union with and then zero to infinity. And then we'll look for whether functions Kong Kate down, which would just be the remaining part of our interval. So negative 0.253 to 0 would be where the function would be con cave down. So now that we have these critical values, we can go ahead and actually uses help us find a nice viewing window for the function. So I think a nice feeling window for this would be negative 3 to 3 negative 10 takes. So now that we have that well here, then we can see much clearer that Oh, it does look like from negative infinity too Infinity. The function I mean Thio about negative heard to about this point here it's increasing Afterwards, it's decreasing and then just decreasing every worlds and we can see from the infinity to maybe about there it looks calm. Keep up. And then after that, it's conch aid down and then on the other side would be conking up his work So you could just do this directly from this graph here. But just to get a nice viewing window, we would do this. But now it tells us to use countless to find these intervals. Exactly. So if we were to set her So first I went ahead and rewrote F climate F double prime as this irrational, there is one fraction because we know if we set the numerator is equal to zero, we can figure out what the intercepts should be. So, looking at this, you might notice that four f prime enough double prime. We only got one point importance. So it seems that we missed something in our graphs. So if I said this equal to zero, um, you could just plug the humor into the quadratic formula, and doing that tells us X is equal to negative eight plus or minus the square root of 61. And you would see that these are approximately equal to negative 0.19 and negative 15.8. So now we were to go ahead and use these to solve for our intervals and so I wouldn't have been solved before handing it just to save a little bit of time. But the function with the increasing on negative eight to negative square root of 61 two negative eight plus square root of 61 and this function is going to be decreasing, decreasing. So this is what have prime of actually strictly less do on negative infinity. Two negative square root of 61 um, negative beats get a big minus. The negatives were 61 and then union that with negative eight plus the square root of 61 20 and then zero two. And so let's go ahead and compare that to what we got the first time or increasing and decreasing. So we said that the function was increasing from negative infinity to negative 0.19 and decreasing from negative 0.19 to 0. And that's your twin thing. So this interval here could actually be broken up Maur for increasing, and we would need one extra interval for decreasing and using, um, the next one setting that equal to zero. So once again, we would use the quadratic formula to figure out where our new miners equal to do and that is going to be X is equal to negative 12 plus or minus the square root of 130. And both of these values are approximately equal to negative. 0.253 and negative 23.7 it right? So let's go help your ladies or Conkey up and concrete down. And I went ahead and solved these beforehand again. So for this to be conch Ava, we need after well, trying to be strictly largest zero. And that occurs on the interval with the intervals negative 12 minus this word of 138 to negative 12 plus the square root of 138 union with zero to infinity and then to figure out where this function is. Kong Kate. Down, down. We want f double prime to be strictly less than zero. And these intervals here going to be negative. Infinity to negative 12 minus square root of 128 union with negative 12. Plus this word of 1 38 two zero. And if we compare that with what we found with estimating our graphs, we can see again that these two intervals here were actually not sufficient. And so if you were to just look at this graph here, and if you zoom really far out, it just really looks like we're getting really close to the ex access on either side. And we would never see this extra bits of information. So that's why it's nice to use thes grafts to maybe get idea of what it looks like. But in the end, we should always go back to the calculus to see if the's viewing Windows were using are actually decent or not.

University of North Texas
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