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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 ( d ) to sketch the graph of $f .$$f(x)=\frac{x^{2}-4}{x^{2}+4}$

A. no vertical asymptote, horizontal asymptote $y=1$B. decreasing: $(-\infty, 0) \quad$ increasing: $(0, \infty)$C. local min: $(0,-1)$D. Intervals of concavity: $x<-\frac{2 \sqrt{3}}{3}$ and $x>\frac{2 \sqrt{3}}{3},$ concave down, $-\frac{2 \sqrt{3}}{3}<$ $x<\frac{2 \sqrt{3}}{3},$ concave up. Inflection points at $\mathrm{x}=\pm \frac{2 \sqrt{3}}{3}$E. SEE GRAPH

Calculus 1 / AB

Chapter 4

APPLICATIONS OF DIFFERENTIATION

Section 3

Derivatives and the Shapes of Graphs

Derivatives

Differentiation

Applications of the Derivative

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in this question. Want to sketch the graph off f? So first, that's for party. We're looking for the sympathetic. So for the horizontal sympathetic, we need to evaluate the falling to limits. So for the 1st 1 when X approaches to negative infinity, fxc FX equals the one and the for the next one. When X approaches to a positive infinity, the limit is also one. So we can prove that we have a horizontal sympathetic Why Costa one when x goes to past minus infinity and that there's no verticals in with Otik in this case because this functions defined it for all X. We won't have anything divided by zero miss for this function. So this is punny for property. We're looking for the increasing and decreasing Terrible. So we need to take the first of the narrative, which is 16 x divided by X squared plus four squared. So we just let, uh, of Prime X equals zero. So we have X equals zero. That means we need to consider referring to intervals from minus infinity to zero from zero to infinity for the first in Tirol. The first degenerative is, um, less than zero. So it's decreasing for a second to roll. The first of the narrative is positive. Say it's increasing. Um, and this result also tells us that Patsy, we have a local, um, Minuteman at X equals zero with a value of zero. It cost a minus one local maximum. We don't have that in the 40. We want to find out. There can cavities. That means we need to take the second narrative, which is, um, 64 miners, 48 x squared, divided by X square pass for Cuba. So just let second of purity. Because zero, we have X equals two pass minus two over route of three. Um so have three intervals from minus infinity to minus two over. Ruutel three from minus two over route off 3 to 2 over. Route off three in the asteroids from two of our route of three to infinity. So for the first in Karol on the second, figurative is less than zero. So it's concave tongue on the second interval, the for the second figurative. It's positive so it can keep up in the last one. It's negative states conclave dumb. Also we can. From here we conclude that there are two inflection points. So you reckon point that X equals two miners to overrule two or three and the at X equals 22 of a route after Now we are ready to graft. Really? A graph this function, my coordinates and the graph our looks like this. So first way labour out the presento sympathetic. Which is why close to one. And we label out to the local Milliman here, minus one. So we have something like this. Be careful that we have twin vaginal points. So, um, the kerf well changes concave ity. Um, at those points, those inflection point size minus two over roots of three to over route off three. And the X intercept will be minus two into. And we have a hurt on those mentality quite close to one. So this is the scheduled A graph.

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