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## Using the first derivative test to find relative (local) extrema

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# Finding relative extrema (first derivative test)

AP Calc: FUN‑4 (EU), FUN‑4.A (LO), FUN‑4.A.2 (EK)

## Video transcript

In the last video we saw
that if a function takes on a minimum or maximum value,
min max value for our function at x equals a, then a
is a critical point. But then we saw that
the other way around isn't necessarily true. x equal
a being a critical point does not necessarily mean
that the function takes on a minimum or maximum
value at that point. So what we're going to
try to do this video is try to come up
with some criteria, especially involving the
derivative of the function around x equals a,
to figure out if it is a minimum or a maximum point. So let's look at what we
saw in the last video. We saw that this
point right over here is where the function
takes on a maximum value. So this critical point in
particular was x naught. What made it a critical point
was that the derivative is 0. You have a critical point where
either the derivative is 0 or the derivative is undefined. So this is a critical point. And let's explore
what the derivative is doing as we approach that point. So in order for this
to be a maximum point, the function is increasing
as we approach it. The function is
increasing is another way of saying that the
slope is positive. The slope is changing
but it stays positive the whole time which means that
the function is increasing. And the slope being
positive is another way of saying that the
derivative is greater than 0 as we approach that point. Now what happens as
we pass that point? Right at that point
the slope is 0. But then as we past
that point, what has to happen in order for
that to be a maximum point? Well the value of the
function has to go down. If the value of the
function is going down, that means the
slope is negative. And that's so
another way of saying that the derivative is negative. So that seems like a
pretty good criteria for identifying whether
a critical point is a maximum point. So let's say that we
have critical point a. We are at a maximum
point if f prime of x switches signs from
positive to negative as we cross x equals a. That's exactly what
happened right over here. Let's make sure it happened at
our other maximum point right over here. So right over here, as
we approach that point the function is increasing. The function increasing means
that the slope is positive. It's a different positive slope. The slope is changing,
it's actually getting more, and
more, and more steep. Or more, and more,
and more positive. But is definitely positive. So it's positive
going into that point. And then it becomes negative
after we cross that point. The slope was undefined
right at the point. But it did switch signs
from positive to negative as we crossed that
critical point. So these both meet our criteria
for being a maximum point. So, so far our criteria
seems pretty good. Now let's make sure that somehow
this point right over here, which we identified in the
last video as a critical point, let's make-- and I think we
called this x0, this was x1, this was x2. So this is x3. Let's make sure that
this doesn't somehow meet the criteria
because we see visually that this is not
a maximum point. So as we approach this,
our slope is negative. And then as we cross it,
our slope is still negative, we're still decreasing. So we haven't switched signs. So this does not meet our
criteria, which is good. Now let's come up with the
criteria for a minimum point. And I think you could see
where this is likely to go. Well we identified in the last
video that this right over here is a minimum point. We can see that. It's a local minimum
just by looking at it. And what's the slope of
doing as we approach it? So the function is
decreasing, the slope is negative as we approach it. f prime of x is less than 0
as we approach that point. And then right
after we cross it-- this wouldn't be a minimum
point if the function were to keep decreasing somehow. The function needs
to increase now. So let me do that same green. So right after
that, the function starts increasing again. f prime of x is greater than 0. So this seems like pretty good
criteria for a minimum point. f prime of x switches signs
from negative to positive as we cross a. If we have some
critical point a, the function takes
on a minimum value at a if the derivative of
our function switches signs from negative to
positive as we cross a, from negative to positive. Now once again, this
point right over here, this critical point x sub 3
does not meet that criteria. We go from negative
to 0 right at that point then go to negative again. So this is not a minimum
or a maximum point.

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