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(a) Find the vertical and horizontal asymptotes.(b) Find the intervals of increase or decrease.(c) Find the local maximum and minimum values.(d) Find the intervals of concavity and the inflection points.(e) Use the information from parts $ (a) - (d) $ to sketch the graph of $ f $.

$ f(x) = 1 + \frac{1}{x} + \frac{1}{x^2} $

$f(x)=1+\frac{1}{x}-\frac{1}{x^{2}}$ has domain $(-\infty, 0) \cup(0, \infty)$(a) $\lim _{x \rightarrow \pm \infty}\left(1+\frac{1}{x}-\frac{1}{x^{2}}\right)=1,$ so $y=1$ is a HA. $\lim _{x \rightarrow 0^{+}}\left(1+\frac{1}{x}-\frac{1}{x^{2}}\right)=\lim _{x \rightarrow 0^{+}}\left(\frac{x^{2}+x-1}{x^{2}}\right)=-\infty$ since$\left(x^{2}+x-1\right) \rightarrow-1$ and $x^{2} \rightarrow 0$ as $x \rightarrow 0^{+}\left[\text {a similar argument can be made for } x \rightarrow 0^{-}\right],$ so $x=0$ is a VA.(b) $f^{\prime}(x)=-\frac{1}{x^{2}}+\frac{2}{x^{3}}=-\frac{1}{x^{3}}(x-2) \cdot f^{\prime}(x)=0 \Leftrightarrow x=2 . f^{\prime}(x)>0 \Leftrightarrow 0<x<2$ and $f^{\prime}(x)<0 \Leftrightarrow x<0$or $x>2 .$ So $f$ is increasing on (0,2) and $f$ is decreasing on $(-\infty, 0)$ and $(2, \infty)$(c) $f$ changes from increasing to decreasing at $x=2,$ so $f(2)=\frac{5}{4}$ is a localmaximum value. There is no local minimum value.(d) $f^{\prime \prime}(x)=\frac{2}{x^{3}}-\frac{6}{x^{4}}=\frac{2}{x^{4}}(x-3) \cdot f^{\prime \prime}(x)=0 \Leftrightarrow x=3 . f^{\prime \prime}(x)>0 \Leftrightarrow$$x>3$ and $f^{\prime \prime}(x)<0 \Leftrightarrow x<0$ or $0<x<3 .$ So $f$ is $\mathrm{CU}$ on $(3, \infty)$ and $f$is $\mathrm{CD}$ on $(-\infty, 0)$ and $(0,3) .$ There is an inflection point at $\left(3, \frac{11}{9}\right)$

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph

Derivatives

Differentiation

Volume

Baylor University

University of Michigan - Ann Arbor

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Okay, so we're being asked to find the vertical and horizontal as hotel of after Beckers. So in order to do that, we look for where there is such a point where his function will have an under finding point and in this case is a little bit more obvious. So at the bottom, you could see that if you plug in zero, we'll get one plus one over zero plus one over one over deal squared, which is still zero, and this is an undefined ah value. So it is un undefined and it will go off to some sort of infinity. We don't know. We got off the positive or negative acquired further evaluation Should I mean ex surgical zoo with the vertical acto, And then to find out whether there is a horizontal jacinto, All we do is we apply limit. So we look at the limit as it goes to positive infinity and goes negative, infinite limited except twenty of one plus one over X class one of X squared. What happened is as Oko do One of her ex will go to zero because as the bottom gets infinitely big against closer and closer to zero. And so this will be zero. And also there is a just noticed There's a slight mistake. I believe, Um, this should be a minus. So let's just fix that. So then this will be minus zero and this is still one. And if you do negative infinity, you also get the same answers. So this is the limit as X goes to positive. Negative. So you have a warzone passenger at quiet Call one for now to find where the increases or decreases we got to find the first derivative cast. I mean, taking the first derivative. So this will come out to be negative one over X squared, plus two over execute. Oh, also, yeah. And that's where we set this equal to zero toe. Bring over one of X square shall be too over execute equal one over x squared and and less cute squared. So multiply both sides by execute and then we'LL have excuse over X squared, which is just X So that means too is equal to X and we also also we have to take in account vertical ascent. Oh, so we have very glass into X equals zero. So we're going to evaluate it. Also at X equals zero when we're doing a signed chart evaluation. So we'll do it at zero and two. We're looking at the sign for a crime. So if you played about Weston zero you got negative numbers you're talking about between gentry. Positive number. You're talking about greater than two. You get negative number, there's going to decrease, increase and decrease. Also remember that zero is a vertical. Ask himto on our original graph because if you look at it when we did the re evaluated earlier, we said that X ical do the vertical ass in tow. So you would think there's a local men here, but there's not s. So there's no local men. We do have a local max occurring to you. The local Max on tactical No, to find the interval of con cavity, we apply the second derivative cast. So we find a double crime and that comes out to be too over. Execute minus six. Rex extra forth without this secrecy. Well, you bring up nice. That's a terrible zero. Sorry about that. And then we add six over extend fourth, both eyes. We got to x cubed equals sticks over actually fourth, Multiply both sides back and forth. So you have extra phones over X cube, which is just again access would be to X equals six and then this will be X equals three and way Do a sign Chart evaluation. Ah, around three. So this will be with three looking at sign of double crime. But he broke It varies Lesson three You get negative number of putting value greater than three A positive number. So you know that it is Khan cave down when his negative infinity to positive three and con came up from three to infinity also, I forgot to do it on the increasing decreasing So we know it is decreasing from negative infinity to zero and from to to infinity And we knows increasing from between Tonto first the left his page. Now we know the con cavity. We can also identify the inflection point to the sign change occurring around ex ical stories. So inflexion point technical story. So now what we do is we have all the information to go on a graph. So I'm gonna go ahead and jar our horses down to ass until right here, so so looked something like this. This is why calls one and we have a local max. It too was crash it Well, we'Ll draw that in. So what the graph will do is we had con cave down for all values Less than three basically so And we cannot go above this line or this is that Is that less truth? So we first we'll have a decreasing So this is the horizontal ascent Open There's a river to class in tow And this is concave down Shake I was going to look something like this I'll go down and then it goes down to infinity Get closer and closer to the X axis and then it comes up We haven't increasing between zero and two, so it goes up Uh uh And okay, they'll be coming. Ah, no slightly go above like with one because we have a local max it too, and then it changes con cavity at three. So we get a conclave up shape, so we have to make a slightly you shape and it will go straight off into the horizontal acid up. So right here's to and this is the graph of F Becks President

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