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In this video we're going to practice finding the local extrema of a function. We want to
see if this function has any local maximums or minimums and observe what those values
are. Here's our function, and actually this is, if you completed the guided example earlier
in this lesson, you've already worked with this function a little bit. The function is
F of X is 3 X cubed plus 4.5 X squared plus 7. We want to find any local maximums or minimums.
How do we start? In general a local maximum or a local minimum value can only occur at
a critical point of the function. Now, not every critical point is going to be a maximum
or minimum, but if there is a local maximum or minimum it's going to happen at a critical
point. The first thing we want to do is find those critical points, and of course to do
that we have to take the derivative. We're going to start by taking the derivative of
this function. This is a nice, simple, friendly polynomial. If I take the derivative here,
if I bring down that 3, that's going to give me 9 X squared plus, bring down the 2, and
that's going to give me 9 X, and then plus 0, so just 9 X squared plus 9 X. We have our
derivative let's find the critical points. Remember the critical points are the points
where the derivative either equals 0, or does not exist. Let's think about if there are
any points where the derivative does not exist. Our derivative in this case is a quadratic,
9 X squared plus 9 X, so that is defined for any X's, for any X value. There are no points
where the derivative does not exist. We can move on to finding places where the derivative
equals 0. To do that I need to solve the equation 0 equals 9 X squared plus 9 X. To solve this,
it's very helpful to factor on the right hand side. I can pull out a 9 X and I'm left with
X plus 1. And I would just set each factor equal to 0. I get solutions of X equals 0
and X equals negative 1. With your critical points you always want to check if they are
in the domain of your original function, but my original function is just a polynomial,
so these are in it's domain, because the domain is all real numbers. I have my 2 critical
points. These critical points are possible extrema of the function, but at this point
I still don't know. I'm going to proceed with the same steps that I would to find the intervals
of increase and decrease at the function. Basically what I want to do is divide the
real number line into chunks based on my critical points and often people do that sort of with
a number line chart that looks something like this. The important points on my number line
are negative 1 and 0, my critical points. I'm going to label them here, and you can
see this sort of divides up the number line into, in this case, three segments. And those
critical points are places where the sign of the derivative could change. Like I would
if I were trying to find the intervals of increase or decrease, I want to test each
of these chunks, or each of these intervals to see if the derivative is positive or negative
in that interval. Let me show you what I mean. My first interval, I'm looking at is here,
this is kind of the first chunk I'm looking at, values that are less than negative 1.
I want to determine if the derivative is positive or negative on that interval. How do I do
that? All I have to do is pick a test point, any test point in that interval and plug that
into the derivative. Let's say I want to test F of negative 2. And I don't really need,
I don't need to know exactly what the value of the derivative is, I just need to know
if it's positive or negative. If I use my factored forms, remember the factored form
of my derivative was 9 X times X plus 1, kind of the short cut is just to think about what's
the sign of each of those factors. For negative 2, 9 X would be negative, and X plus 1 would
be negative, so that's a positive overall. My derivative is positive on that interval.
We can move on to our next interval, which is numbers between negative 1 and zero. Once
again I can just pick a test point. Let's see, I pick negative 1 half. Sorry this should
be F prime of negative 2 and F prime of negative 1 half. Once again I am just going to check
the sign of each of the factors. Let's see. 9 X would be negative, but X plus 1 now would
be positive. Negative 1 half plus 1 is a positive number. This is negative overall. My derivative
is negative on that interval. I just have one more interval to check, greater than 0.
Let's try F prime of 1, let's see if that's positive or negative. 9 X that's going to
give me positive and X plus 1 is going to give me positive. This is a positive overall.
So far our steps have been exactly identical to the steps we used to determine if a function
intervals are increasing or decreasing. If that was what you were asked, at this point,
you could just read off your chart. Right? Where the derivative is positive, if I want
to add an extra layer to my chart here, where the derivative is positive the original function
is increasing and where the derivative is negative the original function is decreasing.
We have a function that is increasing and then at negative 1 that changes to decreasing,
and then at 0 it changes to increasing again. You can write down those intervals if you
needed to. Here what we're concerned with is where the function may have a maximum or
minimum. Remember what the first derivative test tells us. The first derivative test tells
us that if at a critical value the function changes from increasing to decreasing we have
a local maximum there, and that kind of makes sense, right? If I picture in my mind the
function is increasing and then it's decreasing, it makes sense that there would be a local
maximum at that critical point. In this case then we see that we have a local maximum at
X equals negative 1. And the other part, one of the other parts of the first derivative
test says, if at a critical point we change from decreasing to increasing then that critical
point is a minimum. That makes sense too, right? If I change from decreasing to increasing
then I'm going to have a local minimum point. I have both a local maximum and a local minimum
here. Let's record our answers, local maximum and local minimum. We want to be a little
bit careful about the way we report our answers so what the function is asking is, and this
wording may be a little ambiguous but find the local maxima and the minima. It's actually
asking for the local maximum and minimum values. Kind of a good way to report that is F of
negative 1, because then you're telling me where it is, equals, then actually report
that value. F of negative 1, you want to remember to plug those into the original function,
so this going to be 3 times negative 1 cubed plus 4.5 by negative 1 squared plus 7, which
if you work that out gives you 8.5. F of negative 1 equals 8.5 is our local maximum. With the
local minimum that was X equals 0, but what's the actual minimum value? That would be F
of 0 equals 3 times 0 cubed plus 4.5 times 0 squared plus 7, that would be F of 0 equals
7. These are our local maximum and local minimum values. In this case we had one of each but
you can have more than that. One more important thing to note, we didn't see this in this
example, but in theory you could have, let's see, we have a critical point C and it doesn't
change signs, that's something that could happen. In that case that critical value would
not be a local maximum or minimum. One thing you can always do to check yourself is to
quickly look at the graph, so I already have that ready for us. We said there should be
a local maximum at X equals negative 1, with a value of 8.5 and that looks good from here,
and a local minimum at 0, 7, looks good as well. These are the steps you follow to find
local maxima and minima.