[music] Really, why do we have to even bother checking endpoints?
So, let's suppose that I want to find the maximum-minimum values of this function f
of x equals x minus x cubed. That function, f, if I considered on the
whole real line, doesn't achieve a maximum or minimum value.
F of x is really negative if x is really positive.
And conversely, f of x is really positive if x is really negative.
True enough. So, let's ask a better question.
What if I want to maximize and minimize this function on the closed interval for
minus 3 to 3? This is now a continuous function on a
closed bounded interval. So, the extreme value theorem guarantees
that I'll be successful, right? It guarantees that there is some input
which minimizes this function's value and some input which maximizes this function's
value on this interval. I'll begin by listing off the critical
points for this function. So, I'll differentiate this function,
right? And if I differentiate that function, the
derivative of x is 1, and the derivative of minus x cubed is minus 3 x squared.
Alright, so, there's the derivative. I'm looking for places where the
derivative is equal to 0, or the derivative doesn't exist.
Well, this function is differentiable everywhere.
So, there's no critical points where the derivative doesn't exist.
But there are going to be some places where the derivative is equal to 0.
Let's find them now. So, I'm trying to solve this equation.
I'll add 3x squared to both sides, I'm really trying to solve the equation, 1
equals 3x squared. Divide both sides by 3, so now, I'm trying
to solve the equation 1 3rd equals x squared.
Now, I'll take a square root of both sides and I'll find that x is equal to the
square root of a third or maybe negative the square root of a third, so I'll write
x is plus or minus the square root of a third.
These are the two critical points for this function.
Oh, I'm also going to be careful to list the endpoints.
And we'll see why we have to. So, here are the points that I should
check, right? I should check the critical points, and I
just found those two critical points. Negative square root of a third, and
positive square root of a third. And I also want to check the endpoints,
right? And my original question is asking me to
maximize and minimize this function on this interval, which includes the endpoint
minus 3 and the endpoint 3. So, I'm going to include those on the list
of points to check. So, where does this continuous function
achieve its maximum value? And where does this continuous function
achieve its minimum value on the given domain?
So, let's go ahead and just evaluate the function at those four points.
Here are the endpoints, minus 3 and 3, and the critical points plus or minus the
square root of a third. And when I evaluate the function at these
points, what do I find? Well, I find that the largest output
occurs when x is negative 3. And the smallest, the most negative output
occurs when x is equal to 3. At these critical points, the outputs, you
know, in size no bigger than 1, it's about 0.38.
What nightmarish scenario would've ensued had I forgotten to include the endpoints?
Well, here's the danger, right? This is actually what I'm saying.
I'm saying that if x is in this closed interval, then the largest value of the
function occurs when x is equal to minus 3, and the smallest value occurs when x is
equal to 3. If I hadn't included the endpoints in my
chart, I would have been fooled into thinking that the largest output of this
function was 0.38. But on this interval, on this closed
interval between minus 3 and 3, the largest output actually occurs at this
endpoint. The upshot to all of these, is that the
endpoints are, in some sense, critical points.
Standing at an endpoint, you can't wiggle freely, so you really can't differentiate
when you're standing at an endpoint. And a place where you can't differentiate
that's one of the critical points.