24–34. Curve sketching Use the guidelines given in Section 4.4...

Question

24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.
ƒ(x) = ln( x² + 3) / (x -1)

ƒ(x) = ln( x² + 3) / (x -1)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to draw the graph of the function F of X, which is equal to the quantity of X2 + 5 in quantity, divided by the quantity of X minus 2 in quantity, using the information given below. And we're given that F of X is equal to the quantity of X2 minus 4 x minus 5 in quantity, divided by the quantity of X minus 2 in quantity squared, and FX is equal to 18 divided by the quantity of X minus 2 in quantity cubed. But the problem will give an empty graph on which to plot our function. So, in order to graph our function FMX, we need to determine some information about this function. The first thing we need to look at is its domain. And if we look at our function, we see that we have a rational function. And so our function here F of X is going to be defined everywhere except when its denominator is equal to 0. So, if we set our denominator, which is X minus 2, equal to 0 and solve for X, we see that X is going to be equal to 2. So our function F of X is defined everywhere except for X equal to 2. So that means that our domain, it's going to be the open interval from minus infinity to 2. Union with the open interval from 2 to infinity. We can also check for symmetries, because that will help us to graph our function. And so we'll check for even odd symmetries by calculating F of minus X. And this is going to be equal to the quantity of minus X in quantity squared plus 5, and that whole quantity is divided by the quantity of minus X minus 2. And this is going to be equal to the quantity of X2 + 5 in quantity, divided by the quantity of minus X minus 2. And this is not equal to F of X or even symmetry. And this is also not equal to minus F of X. Which would give us odd symmetry, so we have neither even nor odd symmetries here. Now, the next quantities that we can look at are are asymptotes. And here, since we have a fraction, and we found earlier that our denominator is equal to 0, when X is equal to 2, we're going to have a vertical asymptote at X equal to 2. We also need to check for horizontal and oblique asymptotes, and if we look at a function F of X, we see that the degree of the numerator is actually larger than the degree of the denominator. So we're not going to have any horizontal asymptoes. But the degree of the numerator is exactly 1 larger than that of the denominator. And so that means we will have an oblique or slanted aseptote. So to find the equation for that line, we're going to take our denominator, which is X minus 2, and divide it into our numerator of X2 + 5. And we're going to do this by long division. And so X will go into X2 X times. And so when I multiply X by the quantity of X minus 2, I'll get X2 minus 2 X, and I'm gonna subtract this from the quantity of X2 + 5. In doing so, I will get 2 X plus 5. And X will go into 2 x 2 times, so in my question, I'll have plus 2. And when I multiply 2 by the quantity of X minus 2, I will get 2 X. -4. And so must subtract this. From The quantity of 2 X plus 5, and when I do that, I will get a remainder of 9, which I'm not going to be worried about, since for the equation for the oblique asptode, I'm only interested in the quotient. So that means that the equation for the oblique asymptote is going to be Y is equal to X plus 2. So, we can go ahead and draw these asymptotes on our graph. We will have a vertical dash line indicating the vertical asymptote at X equal to 2. And we'll also need to plot some points in order to draw our line Y is equal to x + 2. And so we'll take one point to be Uh, when X is equal to 0, so that means Y is going to be equal to 2. We also look at the case when X is equal to 10, the Y will be equal to 12. And then we'll also look at X equal to -10, and so Y is going to be equal to -8. And so we're just going to connect these points. With a straight desk line. And that's going to be our oblique asymptotes. So, the next quantity that we want to look at are our critical points. And for this, we'll need our first derivative. And if we look at our first derivative, We see again, that we have a rational function. And we call that for critical points. They're going to be the X values which our first derivative is either undefined or equal to zero. Now, we have the same quantity in the denominator, the X minus 2, that we had in our original function F of X. So again, our function F of X is not going to be defined when X is equal to 2. However, X equal to 2 is not in the domain of our function, so we don't need to worry about since we can't have a critical point there. So the other option is to set our first derivative of equal to 0, and our first derro is going to be equal to 0 whenever the numerator is equal to 0. So we're going to set the quantity. Of X2 minus. Or X -5 Equal to 0, and solve this equation for X. Now, on the left hand side of the equation, we can actually factor that quadratic. So this will factor out into X minus 5. As one factor, and that quantity is multiplied by the quantity of X plus 1 or the other factor. Both of those has to be equal to 0. So we'll set each of these factors equal to 0. And solved for X. So we'll have X minus 5 equal to 0, and so that means X is equal to 5, and we'll also have X + 1 equal to 0, and so X is equal to -1. So these are two critical points. So now we need to use these critical points to determine the intervals over which our function is increasing or decreasing, and this will allow us to see whether we have a local maximum or local minimum at these two points. So to do this, we're actually going to make a table, and this table is going to consist of two rows. The first row is going to be the sign of FX, and the second row is going to be the behavior of F of X, whether it's increasing or decreasing. And so we're gonna look at a couple of intervals. The first interval that we're going to want to look at is going from minus infinity to -1. And the sign of F of X is going to be determined entirely by the sign of the numerator, since our denominator, in that case, can never be um negative. And so we can actually use the factor form of our numerator to determine the sign of F of X. So, over this interval, going from minus infinity to -1, the quantity of X minus 5 is always going to be negative. And the quantity of X + 1 is also always going to be negative. So I'll have a negative value multiplied by a negative value, which will give me an overall positive quantity for my first derivative. So I'll put a plus sign in the first row. And recall that if the first derivative is positive over an interval, that means our function F of X is going to be increasing over that interval. So we'll label the second row with an I for increasing. Now, the next interval that I want to look at is going from X equal to -1 to X equal to 2. And again, we'll be looking at the factor form of the numerator to determine the sign of F of X. And so over this interval, The quantity of X minus 5 is still going to be negative, but the quantity of X + 1 is always going to be positive. And so I'll have a negative value multiplied by positive value, which will give me an overall negative sign or my first derivative. And if the first derivative is negative, recall that our function is going to be decreasing. So we'll label that with a D in the second row. Now, the next interval that I want to look at is going from 225. And on this interval, again looking at the factor form of our numerator to determine the sine of F of X. We'll see that the quantity of X minus 5 is again still negative. And the quantity of X + 1 is also still positive on the interval. So again, our overall sign for our first derivative is going to be negative. And that means that our function is decreasing on this interval. And the final interval that we want to look at is going from X equal to 5 to infinity. And again looking at the factor form for the numerator of our first derivative, over this interval, the quantity of X + 5 is going to be positive quantity, and the quantity of X + 1 is also going to be positive. So I'll have a positive value multiplied by a positive value, which will give me an overall positive sign for my first derivative. So that means that our function is increasing on this interval. And if we look at our table here, we see that at X equal to -1, our function is going from increasing to decreasing. So we have a local maximum at X equal to -1, and at X equal to 5, our function is changing from decreasing to increasing, so we have a local minimum at X equals to 5. So, we're actually going to calculate the value of our function at each of these points, since we'll need those to plot them on the graph. So, we'll calculate F of 5. And that's going to be equal to the quantity of 5 squared. Plus 5 in quantity, divided by the quantity of 5 minus 2 in quantity. And when we evaluate this expression, this is going to be equal to 10. And then other quantity that we need to calculate is F of -1, and this is going to be equal to the quantity of the quantity of -1 in quantity squared. Plus 5 in quantity, divided by the quantity of -1 -2. And when we evaluate this expression, We get minus 2. So, the next quantity that we need to look at is our inflection points. And for this, we're going to need to look at our second derivative of X. And possible inflection points occur when our second derivative is equal to 0. And if we look at the case here, our second derivative can never actually be equal to 0, since the numerator is never equal to 0. So that means what we have no inflection points. But we still need to determine the intervals of concavity for our function F of X. So, we'll still be using the second rib to do that. And we're actually going to make a table. And we're going to have the sine of our second derivative, F of X, as one row, and the behavior of F of X as our second row, and we're going to be looking at two intervals corresponding to the intervals for our domain. So, our first interval is going to go from minus infinity to 2. And if we look at our function FXx, the sign of FX is determined entirely by the denominator, since the numerator is always a positive quantity. And so in the denominator we have the quantity of X minus 2. And on the interval going from minus infinity 2, X minus 2 is always a negative quantity, and when I cubit, it still became still remains a negative quantity. So overall, on this interval, our second derivative, FX is negative. And recall that if the second derivative is negative over an interval, that means that that function F of X is concave down. So we'll label that with a D in our second row of our table. Now, the next interval that we're going to look at is going from X equal to 2 to infinity. And again, if we look at the denominator and the quantity of X minus 2, on this interval, the quantity of X minus 2 is positive and therefore, the sign of our second derivative of X X is also positive. And if the second derivative of positive over an info, recall that our function FX is going to be concave up. So we'll label that with a U. So now we've determined the intervals of concavity for a function F of X. And so now we have all the information that we need to actually create our graph. So coming back up to our graph. We're gonna plot our two Xtrema. We'll have one at X equal to 5, and a Y value of 10. And the other point is going to be when X is equal to -1, and Y is going to be equal to -2. We'll plot those points. And so now we're gonna use the fact that our function is always concave down. On the interval going from minus infinity to 2, and when X is less than -1, our function is going to be increasing. And so it's gonna be coming in from infinity close to the Oblique asymptote, reaching a maximum value at X equal to -1. And then decreasing after that approaching. are vertical asymptotes. Then we're gonna be decreasing. As we're coming in from infinity near the vertical asptope. Until he reaches The local minimum at X equal to 5. And then our function is going to Begin increasing Approaching the oblique asymptote as it approaches infinity, and that second part of the function is concave up, which matches what we found on our intervals of concavity. So the graph that we have drawn here on the screen is the answer to this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.
ƒ(x) = ln( x² + 3) / (x -1)

Key Concepts

Domain of a Function

Asymptotes

Critical Points and Derivatives

Watch next

24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.ƒ(x) = ln( x² + 3) / (x -1)

Key Concepts

Domain of a Function

Asymptotes

Critical Points and Derivatives

Watch next

24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.
ƒ(x) = ln( x² + 3) / (x -1)