 In this video, we provide the solution to question number 15 for practice exam number four for math 1210 And this is a curve sketching question in the end We're gonna have to graph the function f of x equals 2x over x square minus one But we can't just graph it. There are some special categories We need to show our work for in order to get full credit on this question So let's look at the first one about domain as this is a rational function The only problem with the domain is what makes the denominator go to zero So we need to solve the equation x square minus one equals zero There's a couple ways you could solve this quadratic equation I'm just gonna do it by factoring it's a difference of squares So we get x minus one times x plus one equals zero We see the domain is gonna be everything except for plus or minus one now It often turns out that the domain has a lot to do with the discontinuities of these functions Not always the case but on this one because we have a rational function We do know that if the denominator goes to zero and the numerator doesn't we have to have a vertical asymptote So there's gonna be vertical asymptotes at the value x equals plus or minus one and if you want to draw those vertical asymptotes right now on the graph you could do so I'm gonna hesitate to do so because maybe I want to draw the scale somewhere other than just one two three four We'll see in a minute. So I'm gonna hesitate to do that. All right What about the intercepts here looking at our function the y intercept is gonna happen when we plug in x equals zero so the y intercept This is just f of zero and if you plug in f u f of zero you're gonna get zero over negative one We see that the the y intercept is gonna be zero in terms of the x intercepts The x intercepts of a rational function occur when the numerator goes to zero So we need to solve the equation 2x equals zero which would then imply that x equals zero So this function is gonna pass through the origin and that's the only intercept x or y How about symmetry well the test for symmetry we need to look at the value f of negative x So we plot we replace each of the x's in the formula with a negative x We get 2 times negative x over negative x squared minus 1 well when you square on negative You just get back the positive version there x square minus 1 you're gonna negative 2x So this is just the original function just times negative 1 so negative f of x this tells me that my function is odd Which means that it'll be symmetric with respect to the origin So that's something we want to incorporate later on into the function That is the graph of the function For the in behavior what we wanted to determine with the in behavior is what happens is we take the limit as x approaches Infinity or negative infinity the very end of the domain So taking the limit as x approaches infinity remember our function was 2x Over x squared minus 1 because this is a rational function It turns out that the limit only depends on the top the dominant terms on the top and bottom So taking the limit as x approaches infinity we get 2x over x squared Which is the same thing as just 2 over x as x approaches infinity and which case then as we allow x to go towards Infinity we see that this is going to look like 2 over infinity which is just zero So that is we have a horizontal asymptote and we're gonna approach zero from above if we take the limit as X approaches negative infinity this calculation basically turns out to be the exact same We're gonna take the limit as x approaches negative infinity of 2 over x We end up with 2 over negative infinity which this will look like zero from the left So we have a horizontal asymptote at the value. Well, just the x-axis And like I said on the right will approach zero from above the positive side and from the left will approach zero from from below All right, so what about Monotonicity can we find the extrema here well since we have the derivative already computed for us and all of the Factorizations already taking place. We are ready to build a sign chart. Maybe let me write this a little bit higher And so what are some critical numbers we should be looking at? Well, you see that the denominator goes to zero at one and negative one like we just explored earlier But negative two times x square plus one that thing never goes to zero x square plus one It doesn't have a real root and so the numbers we're gonna be concerned about are negative one and positive one And what happens to the derivative in that situation f prime in which case you can plug in values like If you want to use test values like x equals zero if you plug zero into this The numerator reminds you it doesn't matter what you choose for x x square plus one's always positive you times that by negative two It's always gonna be negative. So the denominator is always gonna be negative. So it only depends on Excuse me the numerator is always gonna be negative the denominator will switch its signs x squared minus one But I should also mention that you take x squared minus one and you're squaring it What does that mean for us? Well, whatever you take whatever you give for x squared minus one if you square It's gonna be non-negative. So basically what I'm saying is The denominator is always gonna be positive except at the critical numbers the numerator is always gonna be always gonna be negative So what we see on this graph is we're gonna get negative negative negative Always I did that without test points. You could plug in specific test points if you wanted to so what that means for our functions This function is always decreasing That's kind of a curious thing because of the vertical asymptotes the function will always be decreasing But there will be places where it's higher than another spot. It's basically like you fill inside a portal You would fall forever if you kept on falling that portal. That's kind of what happens at the vertical asymptotes All right, so our functions always be decreasing That's important to note on the graph now What about the second derivative the second derivative doesn't really change things up much, right? If you look at the numerator you get x squared plus 3 that thing is always gonna be positive never gonna change the sign You do get a critical number I should say a potential point of inflection at 0 and this x squared minus one cube shows Up again from the quotient rule so in terms of our sign chart this time we want to consider negative one We want to consider positive one and we want to try zero as well And so let's look at what happens at the second derivative when that thing happens Using test values are some other method We can see that it's gonna switch at signs from net at from when you pass negative one the second derivative will be negative Then positive at zero switches from positive to negative and from one if you're gonna go from one to positive again So each of these times it's gonna switch its concavity So what that tells you about the function is this is concave? Downward this is concave upward This is concave downward and this is concave upward This gets us to be a point of inflection at zero Wanted negative one or not points of inflection because they're vertical asymptotes, but concavity can change at Vertical asymptote of course and then what this tells us Yeah, then they were good to go. There's no Maximum or minimum on the graph because it's vertical asymptotes are not extreme points There's no there's no critical number There were no there were no Extreme on this graph if you wanted to you could do both of these tables together do the first and second derivative tests at the Same time, but now we have enough information. We can start graphing our function So some things that were important is that the function goes to the origin So we're gonna include the point x equals zero that definitely was an important point. We need to have our Vertical asymptotes, I'm gonna space this out a little bit Just to make it a little bit easier to draw. There's no reason why X equals one has to be the first tick mark. So we're gonna make this and just label it, right? This is our vertical asymptote x equals one. We have to do the same thing on the other side So let's keep the scale consistent there. So here is our other vertical asymptote Try to draw these straight if you can x equals negative one. I'm also gonna label my my The origin there zero zero because that's our intercept there We also had a horizontal asymptote at the x-axis. So if you want to you kind of draw over it That's perfectly fine And so what remember what we discovered here is that we're gonna approach the asymptote from above on the right We're gonna pass the asymptote below right there and then we are always decreasing So if you start on the left-hand side and you're always decreasing You basically are gonna have to do something like this which this part is decreasing It's also concave downward which we saw early on our table is exactly what should happen when we're left less than negative one So I'm gonna raise those markers there Alright, so then when we go between negative one and zero, what did our charts say? Well, it's gonna be decreasing right here, which is kind of obvious We have to go from we have to go from the Asymptote to the x-intercept there, but wait a second also from negative one to zero The second row is pauses so it should be concave up. So we should expect something like this happening on the graph When you get past the origin, we're still decreasing because the first row is negative But the second row of now is also negative. So it should be concave downward So the origin needs to be a point of inflection. So draw it with that inflection Something like that. And then the last piece to consider is what happens when we're past one Well, when you're past one the function still decreasing, but it's concave upward and we're also approaching our horizontal asymptote from above So the only way to put that together would be something like this And so then this gives us our sketch of the graph of our function f of x equals 2x over x squared minus one We're incorporated all of the stuff about Modesty concavity discontinuities in behavior all of that stuff is incorporated into the picture I'm gonna zoom out a little bit. So you have a chance to try to see everything all at once. I suppose Can't fit everything. Sorry. It's just too much Like there and so thanks for watching everyone