Use the guidelines given in Section 4.4 to make a complete gra...

Question

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) = 3x/(x² + 3)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says graph the following function on its domain. The functions 1st and 2nd derivatives are given. If we're given the function F of X is equal to 8 X divided by the quantity of X2 + 8 in quantity. We're also given FX is equal to the quantity of 64 minus 8 x 2 in quantity, divided by the. quantity of X2 + 8 in quantity squad and F of X is equal to 16 X multiplied by the quantity of X2 minus 24 in quantity, divided by the quantity of X2 + 8 in quantity cubed. We are also given an empty graph on which to plot our function. Now, in order to plot a function, we need to determine several properties of our function F of X. So the first thing we want to look at is its domain. And if we look at our function F of X, it is a fraction. In the denominator, we have a quadratic, and in the numerator we have a linear function, both of which are defined for all real numbers. And if we look at our denominator specifically, we see that it's always a positive quantity, so it's never equal to zero. So therefore, our function F of X is defined on all real numbers. So it's defined its domain is the open interval from minus infinity to infinity. Next, we can look at symmetry, which will help us um to plot our function. And here we can check for even or odds symmetry by calculating F of minus X, and this is going to be equal to 8, multiplied by minus X. Divided by the quantity of minus X in quantity squared plus 8 in quantity, and this is equal to minus 8 X divided by the quantity of X2 + 8. In quantity, and this is equal to minus F of X. So, therefore, we're actually going to have odd symmetry. And so that means that our function is going to be symmetric about the origin. Now, the next quantity that we want to look for are the critical points, and we call that critical points are located whenever our first derivative is either undefined or equal to zero. And if we look at our first der derivative F of X, we see that our denominator is always a positive quantity. And so therefore, our first derivative here is defined for all real numbers. So in order to find critical points, we need to set it equal to 0, and our first derivative here is going to be equal to 0 whenever the numerator is equal to 0. So we'll set 64 minus 8 x 2 equal to 0 and solve for X. So to solve for X, we'll add 8X2 to both sides, so we'll get 64 is equal to 8 x 2. Dividing both sides by 8, we get 8 is equal to x 2, and then taking the square root of both sides, we'll get X is equal to plus minus 2 multiplied by the square of 2. So now we're going to use these critical points determine whether we have a local minimum or maximum at those points, as well as determine our intervals at which our function is increasing and decreasing. To do this, we're actually going to use a table. So this table is going to have two rows. In the first row, we're going to have the sign of our first derivative FX, and in the second row we'll have the the behavior of F of X. And so the first interval that we're going to look at is going from minus infinity to X equal to -2 multiplied by the square root of 2. And if we want to determine the sign of F of X, we just need to look at the numerator, since the denominator in our expression is always going to be positive, so the sign of F of X is always determined by the sign of the numerator. So, we're going to look at the sign of the quantity of 64 minus 8X2, and over this interval, we see that the quantity of 8X2 is always going to be greater than 64. So, that means that 64 minus AX2 is always going to be negative. So, therefore, our first derivative is also going to be negative on this interval. So, we'll label that with a minus sign in our table. And recall that if the first derivative is negative, that means that the function alphabet is decreasing over the symbol. So we'll label that with a D in our table. The next interval that we want to look at is going from X equal to -2 multiplied by the square root of 2 to 2 multiplied by the square root of 2. Now, over this interval, the quantity of 8X2d is always less than 64, so the quantity of 64 minus 8 X2 is going to be greater than 0, so it's going to be positive. That means that our first derivative, FX is also positive on this interval. And recall that if a derivative is if first derivative is positive over an entire interval, that means our function F of X is increasing over that interval. So we'll label that with an I. And the last interval that we want to look at goes from X equal to 2 multiplied by square 2 to infinity. And over this interval, just like with the first interval, the quantity of 8 X2d is always going to be greater than 64, and so 64 minus 8X2 is always going to be negative, and therefore, our F of X is negative over this interval. So that means that our function is decreasing over this interval. So, now we've identified the intervals over which our function is increasing or decreasing, we see that we have a local minimum at X equal to -2 multiplied by the square to 2, and we have a local maximum at X equal to 2 multiplied by the square root of 2. The next quantity that we want to look at are inflection points, and for this we'll need the second derivative. So, if we look at our second derivative function, in order to find uh Inflection points, we need to set it equal to 0. Now, the denominator in our second derivative is always going to be a positive quantity, so it's never equal to 0. So, our second derivative is going to be equal to 0 only when the numerator is equal to 0. So, we're going to set 16 X. Multiplied by the quantity of X2 minus 24 in quantity equal to 0, and solved for X. So this statement is true if either of these two factors are equal to 0, so we'll set each one equal to 0 and solve for X. So we'll have 16 X equal to 0, so that means that X is equal to 0. And for the other factor, we'll have X2 minus 24 is equal to 0, adding 24 to both sides we'll have x 2 is equal to 24, and then taking the square root of both sides, we'll get X is equal to plus or minus 2 multiplied by the square root of 6. So now we can use these possible inflection points to determine the intervals of concavity for a function F of X. So again, we'll use a table, and in this table, we'll have again, two rows. The first row will be the sign of the second derivative of F of X, and the second row is going to be the behavior of F of X, whether it's concave up or concave down. So, the first interval that we're going to look at is going to be going from minus infinity to X equal to -2, multiplied by the square root of 6. And just like with the first derivative, our second derivative sign is determined completely by the numerators, since the denominator is always positive. So, if we look at our numerator, we have two factors. And over this interval, our X values are all negative. So that means our first factor of 16 X is a negative quantity. But if we look at this interval, the quantity of X2d is always going to be greater than 24. So, the next factor, X2 minus 24, is going to be positive. So we'll have a negative value multiplied by a positive value, which gives us an overall negative value for our second derivative. And since our second derivative here is negative, that means that our function F of X is going to be concave down. So we'll label that with a D in our table. Now, the next interval that we want to look at is going from X equal to -2, multiplied by the 2 to 6, to X equal to 0. Again, we need to determine the sign of F of X. And just like the last interval, in this case, X is a negative number, so our first factor in our numerator is going to be negative, but X2d is always going to be less than 24. So the quantity of X2 minus 24 is also going to be negative. So I'll be multiplying two negative values together, which will give me an overall positive quantity. So our second derivative in this case is positive. And that means that our function F of X is going to be concave up on this end, so we'll label that with a U in our table. The next interval is going to go from 0 to 2 multiplied by the square root of 6. And over this interval, our values of X are positive, so the first quantity in our numerator is positive. But X2 is going to be less than 24, and so the quantity of X2 minus 24 is going to be negative. So here I have a positive value multiplied by a negative value, so I'll have an overall negative sign for our second derivative. And so that means that our function F of X is concave down on this interval. And the last interval that we want to look at is going from X equal to 2 multiplied by 26 to infinity. Again, our values for X are positive, so the first term in our numerator is going to be positive, and the value of X2d over this inval is also going to be greater than 24. So X2 minus 24 is also positive. So I'm multiplying two positive values together, which gives me a positive quantity for FX. And so that means that our function is concave up on this interval. And we see that since at every single point that we looked at for possible inflection points that the concavity is changing, all three of the points that we calculated are inflection points. Now, the last quantity that we want to look at are asymptotes, and if we look at our original function F of X, we see that the denominator can never be zero, and so, therefore, we have no vertical asymptotes. So we'll just check for horizontal asymptotes. So for that, we'll calculate the limit as X approaches infinity. Of 8 X divided by the quantity of X2 + 8. And I'm going to rewrite this fraction by dividing both the numerator and denominator by X. So, we'll have the limit. As X approaches infinity. Of 8 divided by. The quantity of X plus 8 divided by X. And so, when we take the limit as X approaches infinity in the denominator of the second term, 8 divided by X is going to approach 0, and that leaves me with just 8 divided by X again. And so that is also going to approach 0. So that means overall this limit is going to be approaching 0. Likewise, if you look at the limit when X approaches minus infinity, Of that same quantity of 8 divided by the quantity of X plus 8 divided by X in quantity, that is also going to be equal to 0 by the same reasoning. So we have a horizontal asymptote at Y equal to 0. So now we have enough information to create our graph. So we're going to have to plot a couple of points. So the first point is going to be when X is equal to -2 multiplied by the square root of 6, which is approximately equal to -4.90. And when we calculate F of minus 4.90, we get approximately -1.22. And so, we're gonna plot that point on the graph. The next point that we're going to look at is when X is equal to -2 multiplied by the square 2, which is approximately equal to -2.83. And we calculate F of -2.83, we get approximately 1.41. So we're going to plot that point on the graph as well. Now, the next point that we need to plot is X equal to 0, and when we calculate F of 0, we get 0, so we'll plot that point on the graph, and now we can just use symmetry to determine the other points. So the next point that we're going to look at, since we have odd symmetry, is when X is equal to 2.83. And our Y value in that case is going to be. 1.41. So we'll plot that point on the graph. And then the next point that we want to look at is X equal to 4.90, and the Y value that we need to plot is 1.22. So we'll plot that point as well. And those are the only points that we need to plot. So now we need to use the fact that as we are coming in from minus infinity, that our function is coming. that it was approaching the X axis as X approaches by its infinity, so we'll be starting off close to the X axis. Our function is going to be decreasing but concave down until it reaches the inflection point, which case it will continue to decrease but now be concave up until it reaches the local minimum. Then it'll continue to be concave up but now increasing until it reaches X equal to 0, which case it switches concavity, so we will now be increasing but concave down. It will continue to increase until it reaches the local maximum. And then it will decrease, continuing to be concave down until it reaches the next inflection point, in which case it's going to decrease but now be concave up, and our function is going to approach the x axis as X approaches infinity. So the graph that we have drawn 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.

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) = 3x/(x² + 3)

Key Concepts

Function Analysis

Graphing Techniques

Use of Graphing Utilities

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