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## Extreme value theorem, global versus local extrema, and critical points

Current time:0:00Total duration:5:51

# Finding critical points

AP Calc: FUN‑1 (EU), FUN‑1.C (LO), FUN‑1.C.1 (EK), FUN‑1.C.2 (EK), FUN‑1.C.3 (EK)

## Video transcript

Let's say that f of x is equal
to x times e to the negative two x squared, and we want to
find any critical numbers for f. I encourage you to pause
this video and think about, can you find any critical numbers of f. I'm assuming you've given a go at it. Let's just remind ourselves
what a critical number is. We would say c is a critical number of f, if and only if. I'll write if with two f's,
short for if and only if, f prime of c is equal to zero
or f prime of c is undefined. If we look for the critical
numbers for f we want to figure out all the places where
the derivative of this with respect to x is either equal
to zero or it is undefined. Let's think about how we can
find the derivative of this. f prime of x is going
to be, well let's see. We're going to have to
apply some combination of the product rule and the chain rule. It's going to be the derivative
with respect to x of x, so it's going to be that, times e to the negative two
x squared plus the derivative with respect to x of e to
the negative two x squared times x. This is just the
product rule right over here. Derivative of the x times e to the negative of two x squared plus
the derivative of e to the negative two x squared
times x, right over here. What is this going to be? Well all of this stuff in magenta, the derivative of x with respect to x, that's just going to be equal to one. This first part is going to be equal to e to the negative two x squared. Now the derivative of e to the negative two x squared over here. I'll do this in this pink color. This part right over here,
that is going to be equal to- We'll just apply the chain rule. Derivative of e to the
negative two x squared with respect to negative two
x squared, well that's just going to be e to the
negative two x squared. We're going to multiply that times the derivative of negative two
x squared with respect to x. That's going to be what, negative four x. Times negative four x, and of course we have this x over here. We have that x over there and let's see, can we simplify it at all? Well obviously both of these terms have an e to the negative two x squared. I'm going to try to
figure out where this is either undefined or where
this is equal to zero. Let's think about this a little bit. If we factor out e to the
negative two x squared, I'll do that in green. We're going to have, this is equal to e to the
negative two x squared times, we have here, one minus four x squared. One minus four x squared. This is the derivative of f. Where would this be
undefined or equal to zero? e to the negative two x squared, this is going to be
defined for any value of x, this part is going to be defined, and this part is also going to
be defined for any value of x. There's no point where this is undefined. Let's think about when this
is going to be equal to zero. The product of these two
expressions equalling zero, e to the negative two x squared, that will never be equal to zero. If you get this exponent to be a really, I guess you could say
very negative number, you will approach zero but you
will never get it to be zero. This part here can't be zero. If the product of two
things are zero at least one of them has to be
zero, so the only way we can get f prime of x
to be equal to zero is when one minus four x
squared is equal to zero. One minus four x squared is equal to zero, let me rewrite that. One minus four x squared is equal to zero, when does that happen? This one we can just solve. Add four x squared to both sides, you get one is equal to four x squared. Divide both sides by four, you get one fourth is equal to x squared. Then what x values is this true at? We just take the plus or minus
square root of both sides and you get x is equal to
plus or minus one half. Negative one half squared is one fourth, positive one half squared is one fourth. If x equals plus or
minus one half f prime, or the derivative, is equal to zero. Let me write it this way. f prime of one half is equal to zero, and you can verify that right over here. And f prime of negative
one half is equal to zero. If someone asks what are
the critical numbers here, they are one half and negative one half.

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