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Current time:0:00Total duration:9:42

AP.CALC:

FUN‑4 (EU)

, FUN‑4.A (LO)

, FUN‑4.A.10 (EK)

, FUN‑4.A.9 (EK)

We've got the function f of x
is equal to x to the third power minus 12x plus 2. And what I want to
do in this video is think about at
what points does my function f take on
minimum or maximum values? And to figure that out I
have to first figure out what are the critical
points for my function f. And then which of
those critical points do we achieve a minimum
or maximum value? And to determine
the critical points we have to find the
derivative of our function because our critical
points are just the point at which our
derivative is either equal to 0 or undefined. So the derivative of this
thing right over here, we're just going to use the
power rule several times, and then I guess you can
call it the constant rule. But the derivative of x to
the third is 3x squared. Derivative of negative
12x is negative 12. And the derivative
of a constant, it doesn't change
with respect to x, so it's just going
to be equal to 0. So we're going to get
a critical point when this thing right over
here, for some value of x is either undefined or 0. Well this thing is defined
for all values of x. So the only places we're
going to find critical points is when this thing
is equal to 0. So let's set it equal to 0. When does 3x squared
minus 12 equal 0? So let's add 12 to both sides. You get 3x squared
is equal to 12. Divide both sides by 3. You get x squared is equal to 4. Well this is going to
happen when x is equal to 2 and x is equal to negative 2. Just to be clear, f of
2, or let me be clear, f prime of 2, you get 3 times 4
minus 12, which is equal to 0. And f prime negative 2 is
also, same exact reason, is also equal to 0. So we can say-- and I'll
switch colors here-- that f has critical
points at x equals 2 and x equals negative 2. Well that's fair enough. But we still don't know
whether the function takes on a minimum values at
those points, maximum values of those points, or neither. To figure that out
we have to figure out whether the derivative changes
signs around these points. So let's actually try
to graph the derivative to think about this. So let's graph. I'll draw an axis
right over here. I'll do it down here
because maybe we can use that information
later on to graph f of x. So let's say this is my x-axis. This is my y-axis. And so we have critical points
at x is equal to positive 2. So it's 1, 2. And x is equal to negative 2,
1, 2. x is equal to negative 2. So what does this
derivative look like if we wanted to graph it? Well we have when x is equal
to 0 for the derivative we're at negative 12. So this is the point y
is equal to negative 12. So this is, we're graphing
y is equal to f prime of x. So it looks something like this. These are obviously the
0's of our derivative. So it has to move up to cross
the x-axis there and over here. So what is the
derivative doing at each of these critical points? Well over here our derivative
is crossing from being positive, we have a positive derivative,
to being a negative derivative. So we're crossing from being
a positive derivative to being a negative derivative,
that was our criteria for a critical point
to be a maximum point. Over here we're crossing
from a negative derivative to a positive
derivative, which is our criteria for a critical
point for the function to have a minimum value
at a critical point. So a minimum. And I just want to make sure
we have the correct intuition. If our function, if some
function is increasing going into some point,
and at that point we actually have
a derivative 0-- the derivative could
also be undefined-- but we have a derivative
of 0 and then the function begins decreasing, that's why
this would be a maximum point. Similarly, if we have a
situation where the function is decreasing going into a point,
the derivative is negative. Remember this is the
graph of the derivative. Let me make this clear. This is the graph of y is equal
to not f of x, but f prime of x. So if we have a situation
we're going into the point, the function has
a negative slope, we see we have a negative slope
here-- so the function might look something like this. And then right at this
point the function is either undefined
or has 0 slope. So in this case it has 0 slope. And then after that point,
let me do it right under it. So going into it we
have a negative slope. And then right over
here we have a 0 slope. Which I could draw it
even better than that. So if we were to imagine
going into it we have a negative slope, right at
that point we have a 0 slope, and then we have
a positive slope. So the function
begins increasing. That's why we say we have a
minimum point right over there. So what I did it
right over here is to try to conceptualize what
the function itself could look like given the derivative,
in this case switching from a positive derivative
to a negative derivative, across that critical
point, or going from a negative derivative
to a positive derivative. That's why this is the
criteria for a maximum point, this is the criteria
for a minimum point. Well with that out of the
way, can we use this intuition that we just talked
about to at least try to sketch the graph of f of x? So let's try to do it. And it's just going
to be a sketch, it's not going to be very exact. But at least it'll
give us a sense of the shape of what
f of x looks like. So my best attempt. So it might not be drawn
completely to scale. So it's my x-axis,
this is my y-axis. We know we have a critical point
at x is equal to positive 2. And we have a critical point
at x is equal to negative 2. We know just from inspection
that the y-intercept right here, if the graph
of y is equal to f of x, when x is 0 f of x is 2. So we're going to
hit right over-- I don't want to
draw this completely to the same scale as the x-axis. So let's say that this
is 2 right over here. So this is where
we're going to cross. This is going to
be our y-intercept. And so we said
already that we have a maximum point at x
is equal to negative 2. So what is f of negative 2?
f of negative 2 is equal to 8 or negative 8,
let me be careful. It's negative 8. And then we're going to have
12 times negative 2, which is negative 24. But then we're going to add it. So we're subtracting
negative 24. So this is plus 24. And then we finally have plus 2. So negative 8 plus 24 plus 2,
so that's going to be negative, let's see, negative 8 plus
24 is 16, plus 2 is 18. So f of negative
2 is equal to 18. And I'm not drawing it
completely to scale. But let's say that this
is 18 right over here. So this is the function. This is the point
negative 2 comma 18. And we know that
it's a maximum point. The derivative going into
that point is negative. The derivative going into that
point is negative-- sorry, the derivative going into
that point is positive. So we are increasing. The slope is positive. And then after we
cross that point, the slope becomes negative. The derivative cross the x-axis,
the slope becomes negative. I actually want to
use that same color. It looks like this. And then, of
course, the graph is going to cross the--
it's going to have a y-intercept,
something like that. And then, as we
approach 2, we are approaching another
critical point. Now what is f of 2? f of 2 is going to be equal
to positive 8 minus 24 plus 2. So this is 10 minus 24, which
is equal to negative 14. So let's say this is the point
negative 14 right over here. Actually I can draw
it a little bit. Let's say this is negative 14. So this is f of 2
right over there. And we saw already that
the slope is negative as we approach it. So our function is
decreasing as we approach it. And then right there
the slope is 0. We figured that
out earlier, that's how we identified it
being a critical point. And then the slope is
increasing after that, the derivative is positive. The slope is increasing. So this is our
sketch of what f of x could look like given that
these are the critical points. And we were able to identify
2 as a minimum point. So this was a minimum value. The function takes on a minimum
value when x is equal to 2. And the function took
on a maximum value when x was equal to negative 2.