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Sketching Graphs Example 3
Hi, everyone!
Welcome back to integralcalc.com.
Today we're going to be doing another graph sketching problem
and the function that we're going to be sketching is f(x) = x^3 - 6x^2 + 9x.
So before we go ahead and get started, let's go ahead and talk about the steps that we're going to take to solve this problem or sketch this graph.
The first thing we'll do is find the derivative f prime of x.
We'll set it equal to zero and solve for x and our answers will be potential critical points of this function.
Critical points tell us where the function is increasing and where it's decreasing.
So we'll plot our critical points in a number line and evaluate between each of them
and that will tell us where our function is increasing and where it's decreasing.
Then we'll take the second derivative which will be f double prime of x,
we'll set that equal to zero and solve for x.
And instead of getting critical points, we'll be getting inflection points.
Inflection points tell us where our graph is concave up and where it's concave down.
So again we'll plug our inflection points in a number line,
evaluate between each of them and that will tell us again where it's concave up and concave down.
So once we've done that,
we will plug all of our critical points and all of our inflection points into our original function, f of x and that will tell us...
that will give us the coordinates of each of those points.
We’ll graph those in our coordinate axis and then use the information that we've gotten
about increasing and decreasing and concave up and concave down to sketch the graph.
So let's go ahead and get started.
The first thing we'll do, like I said is find the derivative so we're looking at f prime of x.
And taking the derivative of our original function term by term, we'll get three x squared minus twelve x plus nine.
So like I said, we'll set this equal to zero and solve for x.
Let's start by factoring a three out of each term,
we'll get x squared minus four x plus three and so it looks like, luckily for us this is very easy to factor.
It's just going to be an x minus three times an x minus one, will give us a minus four x and a positive three.
So if we ...
because this is set as equal to zero, we know that x must equal either three or one because one of these three terms, one, two, three..
One of them has to equal zero inorder for the right side to be equal to the left side of this equation.
So the only way that can be true, obviously, this term here can't be zero because it's always going to be three.
This term can only be zero when x is three and this term can only be zero when x is one.
So our two critical points are three and one.
So now we'll take our critical points like we talked about in the beginning and we'll plot them on a number line.
So our number line is going to run from negative infinity to positive infinity. And we just need to plot our two critical points.
It doesn't have to be proportional;
just as long as they're in the right order with the smallest critical point or the most negative critical point on the left
and then with a positive one on the right.
So as long as one is on the left and three is on the right.
So now we need to evaluate each of these ranges on the number line.
So we've got one range right here another one that runs from one to three and another from three to infinity.
So our critical points have divided our number line into three sections
and we need to evaluate each of them.
inorder to do so, we need to pick a number on each range.
So it doesn't matter what number it is as long as it falls within the range.
So we need any number between negative infinity and one to evaluate that entire range.
So in our case, let's go ahead and pick zero.
We need any number between one and three to evaluate that range so we'll pick two
and then we need any number between three and positive infinity to evaluate this third range so we'll pick four.
So those are the numbers we're going to use to evaluate each of these three ranges
and what we're going to do with those three numbers is plug them in to our derivative equation, f prime of x.
So let's start with zero.
f prime of zero will give us three times zero squared minus twelve times zero plus nine,
which we can see this will be zero, this will be zero so we're going to get nine there.
So it doesn't matter what our answer is.
All that matters is whether or not it's a positive or negative number. In this case, nine is positive.
It is greater than zero, which means that our function on the range negative infinity to one is increasing.
If your answer when you do this is positive, it means your function is increasing.
If it's negative, it means your function is decreasing.
So I like to write both a plus sign to indicate that our function is increasing
as well as drawing this arrow which is a visual indicator to me of increasing.
So now let's evaluate the other two.
So f prime of two equals three times two squared minus twelve times two plus nine.
So let's see…
That's going to be two squared is four so twelve minus twenty four plus nine gives us a negative twelve, negative three.
So that's less than zero.
So our function is decreasing.
So we'll write a negative sign and then we'll draw an arrow that will visually indicate to us that the function is then decreasing on that range.
Finally, we need to evaluate f prime of four to evaluate our third range.
So four squared minus twelve times four plus nine.
So we'll get three times sixteen which is forty eight minus twelve times four is forty eight plus nine.
So we can see that we'll get nine there; which is greater than zero which means our function is increasing on that range.
So again we'll write a positive sign and then draw our arrow that indicates that the function is increasing.
so that's all we need to do for increasing , decreasing.
Now we can get rid of this.
Let’s go ahead and talk about concavity.
So to find out where the function is concave up and where it's concave down, we first need to find inflection points.
And the way that we do that is finding the second derivative of f of x.
So we'll be taking the derivative of our first derivative here. and again, term by term, we'll get six x minus twelve.
So that's our second derivative.
We'll set that equal to zero and we will solve for x.
So I'm going to add twelve to both sides,
I get twelve equals six x and then divide both sides by six, I'll get x equals two.
So that is my only inflection point.
so I'll go ahead and plot that on another number line, negative infinity to positive infinity and here's my inflection point.
Now you can see that this inflection point has divided the number line into two ranges.
One is negative infinity to two, and the other is two to positive infinity.
We need to do the same thing we did with the critical points, find the number on each of those ranges
and then plug it in this time to our second derivative to find out where the function is concave up and where it's concave down.
So we will pick one and three as the numbers at which we'll evaluate.
So we'll do f double prime of one, that gives us six times one minus twelve which is going to be six minus twelve which is a negative six.
That is less than zero which means our function is concave down.
so when you find the inflection points and then you pick your...
the points you will use to evalaute each range and you plug those into the second derivative,
if you get a positive answer, then your function's going to be concave up.
if you get a negative answer, your function's going to be concave down.
So here our function is concave down; which means i'm going to go ahead
and write a negative sign and then to show concavity,
I'm just going to draw a little parabola that is concave down. And that will be my visual indicator.
So f double prime of three,
this time we'll get six times three minus twelve is going to be eighteen minus twelve which is a positive six that's greater than zero.
Which means our function is concave up on this range.
So I'll do a positive sign and then I'll do a concave up parabola as my visual indicator.
So now, our last step is to evaluate each of our critical points and each of our inflection points within our original function.
So we'll be evaluating one, three, and two, our two critical points and our one inflection point inside our original function f of x.
So let's do f of one, our first critical point,
we'll get one cubed which is just one minus six times one which is just six plus nine times one which is nine
and that'll gives us negative five, positive four.
So that means we're going to have one critical popint at one, four.
Now, let's go ahead and evaluate f or three, our second critical point.
We’ll get three cubed which is twenty seven minus...
Let's see...
three squared is nine times six is fifty four so minus a fifty four plus nine times three is twenty zeven, so that's going to be zero.
So we've got a point then at three, zero.
And finally, we'll evaluate our inflection point
so we'll get f of two equals two cubed is eight minus two squared is four times six is twenty four plus eighteen.
And I'm just plugging in each of these numbers into our original function f of x.
So when we simplify this, we'll get eight minus twenty four is a negative sixteen plus eighteen is a positive two
so that means we've got and inflection point at two, two.
Okay. So...
the next thing we're going to do now that we've got all these information is go ahead and sketch our graph.
So let's go ahead and draw some coordinate axis here, straight as possible and we'll start sketching.
So we've got a critical point at one, four, so we'll say that this is one and then one, two, three, four.
So we've got a critical point right about there.
We also have another critical point, three, zero so one, two, three.
We’ve got a critical point right here at three, zero and then we have an inflection point at two, two.
So I'll draw that kind of lightly. So two, two.
So...
our function is increasing.
If we look down here at our little graph that we made, our function is increasing from negative infinity all the way to one.
So that means our function's going to be increasing up to this critical point right here.
Our function is also concave down between negative infinity and two so that means that it's going to look something like this.
This is coming down here and then...
like this...
so we've satisfied between negative infinity and two, the function is concave down.
Looks like that upsiode down parabola.
And also that the function is increasing from negative infinity to one.
We also know that the function is decreasing from one to three.
So this function is going to continue decreasing, right?
From one to three like this and it starts increasing again at three and increases all the way to infinity.
And we can also see that the function is concave up between two and positive infinity
which means that this thing is going to curl back like this.
And I didn't do a perfect job of drawing this at all
but you can see right here is the inflection point where the graph goes from concave up..
or sorry... concave down to concave up.
It’s an upside down parabola and then it becomes a right side up parabola.
So concave down and then concave up and we can see increasing, decreasing and then increasing again.
So that's going to be the graph of our function.
and now all we need to do is write up some supplemental information
that if you have a problem like this on a test, you should always draw the graph and clarify what you found.
So what we do know is let's grab a different color here.
That we have a local maximum here at one, four. We’ve got a local minimum here at three, zero.
And let's see, I guess the other information we've found.
So let's go ahead and say the function is increasing from negative infinity to one.
As well as from three to positive infinity, the function is decreasing between one and three
so decreasing from one to three and then we'll say the function has a local max at one, three.
And the reason it's a local max is because you can see that the function over here is higher than one, three
so there isn't a global max because it increases to infinity.
But if there were a global max, it would be over there but we have a local maximum at one, three
because that’s the highest point in that general area of the graph
So a local max at one, three and then a local minimum at three, zero.
And again, the lowest point in the graph is you know somewhere down here
because the graph decreases to infinity but within the general area of three, zero.
That is the lowest point in the graph so it's a local min.
So that's about it.
That should be sufficient for a final answer.
I hope that video helped you, guys and I will see you in the next one.
Bye!