Graphing functions Use the guidelines of this section to make ...

Question

Graphing functions Use the guidelines of this section to make a complete graph of f.

f(x) = x⁴ + 4x³

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to create a complete graph for the function F of X is equal to X the 4th minus 3 X cubed. Now, in order to create this graph, we're gonna need to determine some information about our function F of X and its behavior. The first thing we're gonna look at is its domain. And here, if we look at our function F of X, we see that we have a polynomial, and a polynomial is defined everywhere. That means our domain is going to be The open interval from minus infinity to infinity. Now, the next thing we want to check for is symmetry, because that could help us actually to create the graph. So, we're gonna check to see if we have even or odd symmetry. So, we're gonna look at F of minus X. And we substitute in minus X for X into our expression for F of X, we we will get. The quantity of minus X in quantity raised to the 4th, -3 multiplied by the quantity of minus X in quantity cubed. Which this is going to be equal to X the 4th, plus 3 X cubed. So this expression is neither. Minus F of X, nor is it equal to F of X. So our function here is neither even nor odd. So, symmetry is not going to help us with our graph. So, the next thing we need to identify are our critical points, and to determine the critical points, we'll need to calculate the first derivative. We'll calculate F of X. Which is equal to the derivative with respect to X. Of the quantity of X4 minus 3 x cubed in quantity. So to take this derivative, we need to apply both our power rule and constant multiple rule. In doing so, this is going to be equal to. Or ex cubed Minus 9 X squad. So, we call that to determine critical points. We need to either find the X values which our first derivative is undefined. In this case, again, we have a polynomial, so it's defined everywhere. And the second condition is when our first derivative is equal to 0. So, we'll set 4 X cubed minus 9 X2. Equal to 0 and solved for X. So I can factor out an X2d out of both of these terms. So we'll have X2, multiplied by the quantity of 4 X minus 9 in quantity is equal to 0. So now I just need to set each one of these factors equal to 0 and solve for X. So for the first factor, we will have X2d is equal to 0, therefore X is equal to 0. And for the second factor, we'll have 4 X minus 9, is equal to 0, and X is going to be equal to 9 divided by 4. So now what we do is use our critical points here to determine the intervals over which our function is increasing or decreasing, and that allows to determine if any of these critical points are local minima, or maxima. So we're actually going to do this by creating a table. So, the first row in this table 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. So here we're gonna be using the first derivative test. So, the first interval that we're going to be looking at is the interval going from minus infinity to 0. And so, I need to choose a point in this interval, so we'll choose X equal to -1, and we'll calculate F of -1. When we do so, we get a value of -13, which is a negative value, so we'll have negative there in the first row. And recall that if our first derivative is negative over an interval, that means our function F of X is decreasing. So we'll label that with a D in our table to represent decreasing. Now, the next interval that I need to look at is the interval from 0 to 9 divided by 4. So again, we'll choose a point in this interval, and so we'll choose X equal to 1. So we'll calculate F of 1. And when I do so, I get -5, which is a negative value, so we'll have again, a negative sign for F of X, and that means that our function F of X is also decreasing on this interval, since our first derivative is negative. And the last interval that we need to look at is going from 9 divided by 4 to infinity. So here we'll choose a point, and the point that we're going to choose is going to be 3. So calculating F of 3, we get a value of 27, which is positive, we'll have a plus sign, and we call that if our first derivative is positive over an interval, that means our function F of X is increasing. So we'll label that with an I in our table. So now what we've done here is we have identified the intervals over which our function is decreasing and increasing, and we actually see that we have a local minimum at 9 divided by 4. So when X is equal to 9 divide by 4, just to the left of that our function is decreasing, and just to the right of that, our function is increasing, which is what we see when we have a local minimum. So now, what we need to do is determine inflection points in concavity, and for this we'll need to actually calculate our second derivative. So we'll calculate F of X. Which is going to be equal to the derivative with respect to X of our first derivative, which is the quantity of 4 X cubed. Minus 9 X squad in quantity. So again, applying our constant multiple and power rules. This is going to be equal to 12 X squad. Minus 18 X. Now, in order to determine possible inflection points, we need to set our second derivative here equal to 0 and solve for X. So we'll have 12 X squared. -18 X. is equal to 0. So we can factor out a 6 X out of both of these terms. We'll have 6 X multiplied by the quantity of 2 X minus 3 in quantity, and that's gonna be equal to 0. So, again, we'll set each one of these factors equal to 0 and solve for X. So we'll have 6 X is equal to 0, therefore X is equal to 0. And for the second factor, we'll have 2 X minus 3 is equal to 0, and therefore X is going to be equal to 3 divided by 2. So, in order to determine whether these points are inflection points, we'll have to look at the concavity of our function F of X. To do this, we'll use the second derivative test. And so we're gonna again make a table where one row, the first row, is going to be the sign of F of X, and then the second row is going to be the behavior of F of X. So the first interval that we're going to look at is going to be the interval, going from minus infinity. 20. And here we'll choose a point in this interval, so we'll choose X equal to -1, and we'll calculate the value for our second derivative. So, we're looking at F of -1, and when I calculate that value. I get a value of 30, which is positive. And recall that if our second derivative is positive over an interval, that our function F of X is going to be concave up, so we'll label that with a U in our table. The next thing that we need to look at is going to be from 0. To 3 divided by 2. And again, we'll need to choose a point in this interval, so we'll choose the point X equal to 1, and we'll calculate F of 1. And when I calculate that value, it turns out to be -6, which is negative, and recall that if our second derivative is negative, that means our function F of X is concave down. So we'll label that with a D in our table. And finally, the last interval that we need to look at is going from 3 divided by 2 to infinity. And here we'll choose a point in this interval, so we'll choose X equal to 2. And when we calculate F of 2, We get 24, which is a positive number. And so that means that our function F of X is concave up on this interval. So that means that we do have inflection points at X equal to 0 and X equal to 3 divided by 2, since the concavity is changing from up to down, when X is equal to 0 and from down to up when X is equal to 3 divided by 2. So now we have enough information to begin creating our graph. So, we'll need a graph area. Over which to graph, and so we've copied that on to the screen. And so, we'll need to identify some points for this graph. So the first point we're gonna look at is X equals 0, and if we calculate F of 0, we get 0, so the first point is going to be at 00. The next point that we're gonna look at is the 2nd inflection point, which is X is equal to 3 divided by 2. And when we calculate F of 3 divided by 2, we get minus 81 divided by 16. And so we'll plot that point on the graph as well. And then the next point that we're gonna need to look at is X is equal to 9 divided by 4. So, when we calculate F of 9 divided by 4, we get approximately minus 8.54. So we'll plot that point on the graph as well. And then finally, we'll just add one more point plot, so we'll plot X when X is equal to 3, and so F of 3 is actually equal to 0, so it'll just give us another point with which to create our curve. So now we want to do is use the fact that For points less than X equal to 0, I have a concave up function. And so, I'll draw a function that is concave up. And then we'll have an inflection point at X equals to 0, and my function is going to be concave down to the point of X equals to 3 divided by 2. Then again, we'll have a concave up curve, but we'll have a local minimum at X equal to 9 divided by 4. And then we'll just have our function increasing after that. So, the graph that we have drawn on the screen here is the answer to this question. So I hope that this explanation has been useful, and I'll see you in the next video.

Graphing functions Use the guidelines of this section to make a complete graph of f.

f(x) = x⁴ + 4x³

Key Concepts

Polynomial Functions

Critical Points and Extrema

End Behavior of Functions

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