Maxima and Minima on a Closed Range


Uploaded by TheIntegralCALC on 04.05.2011

Transcript:
Maxima and Minima on a Closed Range Example 1
Hi, everyone.
Welcome back to integralcalc.com.
Today we're going to be evaluating a function for maxima and minima on a closed range.
So we've been given the funtion f of x equals sine squared of x.
And we've been asked to find its extrema points, it's maxima and minima, on the range x equals zero to x equas three.
So between those two lines on that range, we're looking for maxima and minima of the graph.
So inorder to find them, the first thing we'll need to do is find the derivative, f prime of x, the derivative of our original function.
So f prime of x.
To find the derivative, the first thing I would like to do is convert this equation, this function into the following form.
I'm just going to move the exponent outside of the function. I haven't changed the function at all.
I’ve just rewritten it to a different form and this form, you should know,
is technically not a correct way or the most correct way to write sine squared of x.
You should write the way it's given to us in the original problem.
But for me, it's easier to visualize taking the derivative of sine of x squared.
So I temporarily moved the exponent outside and then I'll bring it back in in a step soon.
So to find the derivative of sine of x squared, we're going to need chain rule.
And the reason is because remember with chain rule, we’ve got an outside function which is basically, x squared, right?
We ignore sine of x, just treat it like a variable x and we've got x squared or something squared.
That’s our outside function. The inside function is just sine of x, right?
So if you're having trouble with chain rule, I've got a whole section about that on my website that you can check out.
But basically, what we do is we take the derivative of the outside function first, leaving the inside function completely alone.
We don't change it at all. We leave it seating there exactly as it is.
And then once that's done, we multiply by the derivative of the inside function.
So in other words, to take the derivative of this using chain rule, we'll take the derivative of the outside function first.
So if we ignore sine of x and just treat it as like a variable,
just pretend that it's x, we would bring this two out in front, right?
It’s just like the derivative of x squared.
The derivative of x squared would be two x.
So the derivative of sine of x squared is going to be two sine of x, right?
So we leave sine of x completely alone. But then we have to multiply by the derivative of the inside function.
And the inside function is sine of x.
Multiplying by the derivative of sine of x means we're multiplying by cosine of x.
So that means that our derivative function is just going to simplify to two times sine of x cosine of x. So that's it.
Now, in order to continue with our maxima-minima problem, we're going to set the derivative function equal to zero and solve for x.
So to solve for x, let's first go ahead and just divide by two to get that out of the way so we'll get zero equals sine of x times cosine of x.
Now we need to realize that the only way that our right hand side here is going to be equal to zero
and therefore, satisfied the condition we've got over here equal to zero.
That means that either sine of x has to be equal to zero and/or cosine of x has to be equal to zero.
So we can actually solve for these separately.
We can find... we can solve sine of x equals zero first
and then solve cosine of x equal to zero and all of our answers combined will give us our critical points.
So sine of x is only equal to zero at x equals zero and pi.
We know that from the unit circle but your calculator will also do it.
So x equals zero and pi and then we set cosine of x equal to zero and cosine of x
is only equal to zero at pi over two and three pi over two.
So these are our critical points.
And if we order them, we'll get x equals...
we're going to order them from smallest to largest.
We’ll get x equals zero, pi over two, pi, and three pi over two.
We just ordered them from smallest to largest.
So as with any maxima-minima problem on a closed range, we find our critical points like this.
But then before we can go further, we should evaluate our critical points to make sure that they fall within our range.
Remember, that we're going to be evaluating on the range zero to three.
That was given to us on our original problem right here; which means we only care about critical points that fall within that range.
There might be others of this function that are outside that range but they don't matter to us because we're only looking at zero to three.
So first of all, we see that we've got end points zero and three,
which means zero is not going to be a critical point that we look at because we've already got that as our end point.
It has to be within the range inside of it.
And we know that pi, right?
pi is three point one four something, which is going to be outside our...
the righthand limit of our range right?
Our range ends at three and this pi is three point one four which means that's gone too.
And since three pi over two is greater than pi, that's also gone.
Which means our only critical point is pi over two.
So we’ve just reduced the number of critical points we have to look at from four to one which is awesome.
So we can go ahead...
to save ourselves some space and get rid of all this, our only critical point remember is pi over two.
So we'll get rid of that.
Now that we have our critical point, we have to plot it on a number line.
So were just going to go ahead and put pi over two, our only critical point on the number line.
The endpoints of our number line are going to be the endpoints of the range given to us in our original problem, zero and three.
So what we've done essentially, is divide our range, zero to three into two sections.
One that runs from zero to pi over two, and the other one that runs from pi over two to three.
And if we look at pi over two, if we evaluate pi over two on our calculator, it's about one point five seven.
So that is helpful to know because we're going to need to pick a number to act as a representative for each of these ranges.
So we need to pick a number that's between zero and one point five seven.
So I'm going to go ahead and pick one.
And then we need to pick a number that's between one point five seven and three so I'm going to pick two.
Now that we have these numbers, we're going to take them and plug them into our derivative function.
The answers that we get will tell us whether or not our function is increasing or decreasing over each of these ranges,
zero to pi over two and pi over two to three.
So we're going to evaluate the derivative function first at one.
We’ll do this one first. So f prime of one.
Remember that this here is our simplified derivative function.
So we'll get two sine of one times cosine of one.
And we don't have to be real pretty about our answer.
All we need to do is plug into our calculator and see what we get.
So when I plug this into my calculator, I get approximately point nine zero.
And it doesn't matter at all what the actual answer is here.
All that matters is whether or not it's a positive or negative number.
So obviously point nine zero is a positive number.
It’s greater than zero which means that our function is increasing over the entire range, zero to pi over two. Increasing.
So essentially, when we plugged in these numbers here, right?
We’ve said they're going to act as representatives for the entire range.
So one is a reperesentative number for the entire range zero to pi over two.
And when we plug one into our derivative function, we'll either get a positive or negative number.
If we get a positive number, it means that our function is increasing over the entire range zero to pi over two.
If we get a negative number, it means that the function is decreasing on that range.
So when we determine whether or not it's increasing or decreasing, I like to draw this big arrow here.
That’s a visual indicator to me that the function is increasing.
I’ll draw it up into the left...
or sorry, up into the right.
And I'd also like to write a positive sign here to indicate that it's increasing.
So now we'll evaluate f prime of two to see whether or not the function is increasing or decreasing
over the range pi over two to three and we'll get two times sine of two times cosine of two.
And when we use our calculator to find the answer to that,
we'll get two sine of two times cosine of two gives us a negative point seven five, approximately.
And these are ugly decimals.
I’m just rounding to two places
At negative point seven five and again it doesn't matter.
All that matters is that it's a negative number, which means our function is decreasing over that range.
So I draw my decreasing arrow and I write my negative sign in there.
So we can go ahead and get rid of all these. Because we don't need it.
So we know where our function is increasing and where it's decreasing.
And now as you can see, right?
Our function could look something like this, right? It’s increasing and then decreasing.
Which... and this is again why I like to draw these arrows.
This gives us a clear visual indicator that that critical point, pi over two is going to be a maximum for our function because you can see it.
The graph goes like this, which means we've got a maximum.
That’s the highest point on that orange line right there, that orange curve of this graph.
So given that, the last thing we need to do is evaluate our critical point and our two endpoints in our original function.
So we're going to evaluate zero, pi over two and three into our original function.
And the answers we get will tell us where our global and local extrema are.
So we'll go ahead and first start with zero.
So well plug in zero to our original function, which is going to be sine squared of zero.
So if we do sine of zero, that's zero.
So that means sine squared of zero is going to be zero. Now we'll evaluate pi over two.
So f of pi over two equals sine squared of pi over two
and pi over two on the unit circle is going to be one so we'll get one squared which is just one.
And then we'll evaluate three so f of three gives us sine squared of three.
And I don't know what sine of three is.
So.. then we have to square that.
So it's approximately...
sine of three squared is point zero one nine nine, approximately. We get a long decimal. But that's good enough.
So what this tells us... what we just did here, we took x-coordinates of zero, pi over two and three
and we found their corresponding y-coordinates right here, zero, one and point zero one nine nine.
So we need to look at these three right here... one, two, and three...
And we need to determine which of these is the largest number.
Well, clearly, one right here is the largest, which means and that makes sense because that's at pi over two and that looks like a maximum to us.
So that means...
since one is larger than both zero and point zero one nine nine, that we have a global maximum at the point pi over two, one.
And now if we look for the smallest number, right, which is going to be zero because that is smaller than both one and point zero one nine nine,
that means we're going to have a global minimum at the point zero, zero.
So we have that and if we look at f of three...
we could just leave it like this, global max and min.
But if you've been asked to find local extrema as well,
our endpoints zero and three here, this is our endpoints on our graph.
f three, you notice that the function is... let me grab...
you notice that the function is decreasing like this right?
Which means that within this range, within this area of the graph,
the endpoint here is the lowest point in that area because the graph is coming down,
it's decreasing, toward our endpoint of three which means that right at three, that's going to be the lowest point in that area of the graph.
Which means that we can say that this graph, given these endpoints has a local minimum at the point three
and then instead of writing our gross decimal point, zero one nine nine,
we'll do three and then sine squared of three which is a cleaner way to write the coordinates there.
So again, the reason it's a local min is because you can only have only one global minimum.
Global minimum means it's the most, the lowest point in the entire graph.
Local minimum means that in that area roughly, it's a minimum point but for the graph as a whole, it's not the absolute minimum.
So we've just got a local minimum in that area, three comma sine squared of three.
So that's it.
I hope that video helped you, guys and I will see you in the next one.
Bye!