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Question with last attempt is displayed for your review only16r For the above rational function f( x ) 1.522 lowest Previewidentify its three inflection points_midd...

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Question with last attempt is displayed for your review only16r For the above rational function f( x ) 1.522 lowest Previewidentify its three inflection points_middle Preview highest Preview Enter mathematical expression Wdlee Message instructor about this question

Question with last attempt is displayed for your review only 16r For the above rational function f( x ) 1.522 lowest Preview identify its three inflection points_ middle Preview highest Preview Enter mathematical expression Wdlee Message instructor about this question



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In Exercises 5–24, analyze and sketch a graph of the function. Label any intercepts, relative extrema, points of inflection, and asymptotes. Use a graphing utility to verify your results.
$$
y=(x+1)^{2}-3(x+1)^{2 / 3}
$$

Just judging from the graph here, if we were to plot this and look at where the vertical awesome tote is, it looks like it's pretty close to negative one. However, if you plugged in negative one, it would still have a value. And so we know that it's somewhere around that. Then Part B is a way to I'm sorry. Negative one or B is a way to verify exactly where that value is by setting Q equals zero and looking for the roots, and that ends up being negative 0.9 to six 073 And so this is where the value the vertical Assam tow it happens. And where this function is undefined or discontinue s, we could say.

Okay, what we're going to do is talk about curve sketching. So we want to develop a nice graph of the function F of X is equal to x minus three over X. And so the first thing we need to do is actually talk about the domain. Um And then you notice that typically the domain is all real numbers or from negative infinity to positive infinity. Unless the variable is in the denominator. And we know our denominator cannot go to zero. So that is actually going to also define what is called a vertical ascent. Hope so we have a vertical assam tope at X equal to zero, which means our domain goes from negative infinity to zero, union with 02 infinity, it cannot equal zero. And I'm actually going to simplify this. So I'm actually going to rewrite this as one minus three over X as my function. Then I also noticed that if I pull from pre calculus, my horizontal Centobie I actually have a horizontal isotope. So any time my highest exponent, my numerator is exactly equal to the highest exponent in the denominator. Then my horizontal Alison tope is leading coefficient over leading coefficient. So that is actually going to be equal 21 I can also use my in behaviour test. So my in behavior tests is actually the limit as X goes to positive and negative infinity of one minus three over X. And so you notice that will just actually equal one. So both of my ends will actually be approaching that horizontal as um toque So now we're actually going to do, let's actually do X intercepts. So where does it cross the X axis? So if I let y equal or f of X equals zero then I get three. As my ex intercept. My Y intercept is when X equals zero and you know that cannot be. So we have no y intercept. Now let's go ahead and do our 1st and 2nd derivatives. So I'm gonna take my first derivative. And when I take my first derivative I actually get um three over X squared. And so then I said um I have no critical numbers but I do know X cannot equal zero. Then if I do my second derivative I get negative six over X cubed. And once again I have no critical numbers based off of that. But I do know the X once again cannot equal zero. So let's go ahead. I'm going to go ahead since those are the same. I'm going to divide from negative infinity 20 and zero to positive infinity. So if I pick a test point negative one and one I'm going to evaluate my first derivative at negative one and I that is positive. So I know my function is increasing. If I evaluate it at positive one, I still get positive which also means my function is increasing. So now if I evaluate my second derivative at negative one I get positive. Which means my function is concave uh huh. And if I'd evaluate it, my second derivative at a positive one, i it is negative which means is concave down. So let's put this all together. So I do know I have a vertical athm tope at zero. So we're gonna go ahead and put that little boundary line in at one. I have we have a horizontal assam Tope and it crosses the X axis at 30 So I know from negative infinity. So way out here two zero X. Equal to zero. My functions increasing in concave up. So there is that portion of the graph and then from zero to positive infinity. My function once again is increasing, but it's concave down and it has to cross the X axis at three and notice that both of my ends are approaching that horizontal assam Toby. Now, of course, you can always verify this graph by graphing in a graphing calculator just to just to verify that we did this correctly.

Okay, what we want to do is step through the process of being able to develop a pretty good graph of the function F of X is equal to X cubed over X squared minus nine. So the first step is to talk about the domain and a typically for functions at domain is horrible numbers except if the variables in the denominator we know that we cannot divide by zero. So those are actually also going to be my vertical assam hopes so X cannot equal plus and minus three. So that means those are my for the class and hopes, which also means those are also my domain restrictions. So my domain is going to go from negative infinity two negative three, Union with native three to positive three. Union with three tim fendi. So those who can be actually be my vertical isotopes horizontal santo. We know from pre calculus that if my highest exponent in my numerator is one more than my highest exponent in my denominator, I do not have a horizontal Assen tobe. But I actually have was called a slant out some tope. And so if we actually do long division, we actually get a slant awesome tope equal to why equal two X. So we're just going to do long division on this and find our slant as santo is why equal to X. It is always going to be the quotient, ignoring the remainder. And then um let's go ahead and do our X intercepts. So our X intercept is when Y is equal to zero. So that is going to be 00 And then our Y intercept is when X is equal to zero, which is also 00 Okay, so then let's go ahead and do our first derivative. So f prime of X. And so this is actually going to be a quotient rule. So it's the derivative of our numerator times are denominator minus our numerator times the derivative of our denominator, which is two X. All over our denominator squared. And of course we're in simplify and when we simplify it reduces down to reduces down to a X squared times an X squared minus 27. All over X squared minus nine. Cute. And so for our critical numbers, we set our derivative equal to zero and so are critical numbers are X equal to zero and plus and minus three. Route three. And of course X cannot equal plus and minus three here as well. Okay. Um And then let's go ahead and do our second derivative as well. So our second derivative once again is actually going to be a quotient rule. And so we're gonna go ahead and do that. And it's a pretty ugly quotient rule. So we'll go ahead. Um Here, I went ahead and distributed the X squared, So that way I didn't have a product role and a quote role at the same time. And so this will be four X cubed minus 54 X times are denominator minus our numerator, which is going to be the X to the fourth minus 27 X squared times are derivative of our denominator, which will be a two times X squared. That should still be a square down here minus nine times a two X. And now I will square that denominator again. So X squared minus nine to the fourth. Okay. And then when I do a bunch of simplification, we actually get 18 X times X squared plus nine. All over the X squared minus nine. And that's that is cute. Okay. And so where does our second derivative equals zero for those critical numbers? And that is when X is equal to zero and X once again cannot be plus or minus three. Okay, so let's go ahead and kind of get our table all set up. So we're gonna divide our table. So we're gonna go from negative infinity to negative three. Route three negative three. Route three two negative three negative 3 to 0 zero to positive three 323 Route three. And in three route three to infinity. Okay, so let's pick a test point in each of those. So negative six, negative four negative two 24 and six. Okay, so let's evaluate our first derivative. So I may go back and do my first derivative at each of those. So if it's negative six then that will actually be positive. If I do a negative for up here then this will actually be negative. If I do a negative to appear, that will also be negative. If I do a positive too, that is going to be a negative. Cause I'm really looking at this term right here and if I do a positive four, that will be negative and a positive six will make that positive. Okay, so let's think about what our function is doing. So he will be increasing decreasing. So that means I will have a relative max here. This is a vertical isotopes. So we're gonna stay decreasing, decreasing, decreasing. And then I go from decreasing to increasing, which means I have a relative minimum. So I have relative max at negative through three. So if I substitute that back into my original function, I get negative nine route 3/2. And then I have a relative men at positive screw three times a scroll 23 And I when I substitute that back into my original function, I get a positive nine scroll 2. 3/2. Okay, now let's do our second derivative. Okay, so I know for my second derivative I don't have to worry about these. So I'm really just going to go from negative infinity two negative three. So it may combine these two. And so it doesn't matter if I do a negative four. If I do a negative for here, then this is for this whole region, it's going to be negative. Which means my function is concave down. And then from negative 3 to 0 he's going to be positive, which means he the function is concave up and from 0 to 3 he's he's negative, which means he's going to be concave down. And then I'm combining these two right here as well. And when I evaluate him at positive six, my second derivative is positive. Which means my function is concave. Uh huh. Okay. So let's see since right in here I'm changing direction of calm cavity. I actually have a point of inflection at zero X. Equals zero. Evaluate my my original function at zero. I actually get zero. So let's put this all together. Let's see if we can do this. Um So we might have to go back and forth. So we know we have an X. Y X intercept, Y intercept and a point of inflection at 00 So let's get that point mark. We also know that we have vertical isotopes. We remember correctly at positive and negative three. And we have a slant assam. Tope at why equal two X. So it's going to be a diagonal line. Why equal to X. Okay. So if I actually put in my relative extremists in my calculator um this value is about negative 5.2. And so this is a positive 5.2. This is a negative 7.8. That's a positive 7.8. So I have a relative men at about 5.2 and 7.8. So it's going to be somewhere up here and we know that down here is kind of the same thing. So that's gonna be a relative max increasing to that point and in decreasing and I'm calm cave down, that entire thing. So I know this is going to be something like that. This will be something like this because it's going to be concave up over that entire region. Now let's talk about that middle part. So I know that concave up and that middle part is I am actually decreasing lips so I'm actually decreasing. But I'm concave up and then I'm decreasing but I'm calling cave down and that is what that should look like. And of course you can always verify by actually graphing on your calculator

Okay, we're going to walk through the process of being able to do a nice graph of Y equal two negative one third X cubed minus three X plus two. Um This is called curve sketching, so we're gonna be doing quite a bit. Um And the first thing we need to do is one you already kind of know from previous middle class is what cubic function kind of looks like, I think we're going to be. Um And we also know that the domain for a polynomial function is negative infinity to positive infinity. Um And then we're also going to talk about kind of how to maybe determine some X and Y intercepts. And so let's do ahead and do um the Y intercept. And that is when X zero and so when X zero, this is here at negative two thirds. My ex intercept kind of don't want to I don't really want to factor that thing. Um So let's see if we can do this. What let's see if we can do this kind of um We know that's going to be y called zero. Probably will have two of them. If not three of them, maybe two. Um And so let me do that's not zero. If I could do a one uh here that will that will work. Um And then let's see if we can maybe do a second one. If it's a negative too. Let's see if negative to zero work. So that would be a negative eight plus A six minus, yep. That works as well. Um Probably would have been easier to kind of factor um that as well. Um And probably more efficient to factor instead of trying to guess, especially if it's not going to be a nice hole number. Um And then let's do our first derivative or no less. You are in behavior so lets you are in behavior. So the limit as X goes to negative infinity of our function. So I believe that is actually going to be equal to positive infinity because this will be negative times a negative, which will give me a positive. And then the limit as X goes to positive infinity, we're going to have positive and this will be a negative. Okay? Um and now let's do our first derivative. So this will be negative one third. This will be three X squared minus a three. And let's go ahead and said that equal 20 And so if that is equal to zero, then I have X is equal to plus and minus one second derivative. We have a negative one third times a six X. So this is going to be actually equal to a negative two X. And so where does my second derivative equal zero? And that is going to be when X equals zero. Okay so I think we have all that set and so now let's go ahead and set up our intervals. And so we're gonna go from negative infinity two negative one negative 1 to 0 0 to 1 and then one to infinity. And so let's pick some test points. And let's evaluate our first derivative at negative two. And so that will be positive and that will be negative. So this will be negative uh negative one half. This will be negative and positive at positive one half. This will actually be negative. So that will be positive. So it's believe that's gonna be positive and probably needed to know that because that is not part of zero is not part of that anyways. Um And at to this is gonna be negative. Okay. So I'm going to be decreasing increasing increasing again and in decreasing. And so I know I'm going to have a relative minimum here at negative one. And so when I substitute that back into the original function then I know that that is going to be negative four thirds. And then I'm going to have a relative max at positive one. And when I substitute that back into my original function, that is actually going to be equal to zero. Okay, um and now my second derivative from negative infinity. So negative two. That's gonna be positive. This will be positive of course, because that whole thing is going to be let me do that. So I don't have a break here. So I'm just going to do positive here and then here in this region right here then if that is that's going to be negative. So we have, our function is paragraph is concave up and then concave down. And so I have a point of inflection at zero and at zero we are at negative two thirds. Right? So let's go, let's see if we can put all this together. So I know I have at zero, negative two thirds is my y intercept. And it's also my point of inflection. I have a relative max at 10 and then I have a relative minimum at negative one and a little over this negative one. Um and then I also have negative 20 is actually another X intercept. So I'm decreasing, I'm decreasing all the way to this minimum and then I'm increasing to hear and then decreasing again. And then I also am concave up from zero to this point right here and then I go to concho down. So I believe that is what it should look like. And of course I would label on my points and then we can always verify using a graphing calculator


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