Main content

### Course: AP®︎/College Calculus AB > Unit 5

Lesson 2: Extreme value theorem, global versus local extrema, and critical points# Finding critical points

Sal finds the critical points of f(x)=xe^(-2x²). Created by Sal Khan.

## Want to join the conversation?

- What is the difference between critical numbers and critical points ?

Are they the same ?(67 votes)- Sometimes they're loosely used interchangeably, but typically when you're asked to identify a critical point, you're expected to provide the coordinates of the point (both x and y), and a critical number is just the x value: the place on the number line where f' is zero or undefined.(160 votes)

- How do we know which point is going to be the maximum or minimum point?(17 votes)
- Assuming you have figured out what the critical points are, you can just take any one convenient number between each two neighbouring critical points and
**evaluate the derivative function f'(x**) at those points that you have chosen.

Then you look at every critical point and check—using your new data—if the derivative is negative before it but turns positive after it (makes it a**minimum**point)*or*is positive before but turns negative (**maximum**)*or*doesn't change sign, in which case you don't care about that critical point.

[**Justification**] That works, because where the derivative is positive, the function is increasing and where it is negative, the function is decreasing. If it steadily increases up to some point and then starts dropping down, that's obviously a maximum. The opposite case corresponds to a minimum.(38 votes)

- Are all global maximums and minimums also local?(17 votes)
- There is only one global maximum (and one global minimum) but there can be more than one local maximum or minimum.(0 votes)

- at the juncture where we analyze which of the terms e^(-2x^2)* (1-4(x^2)) is 0, it is mentioned that the 'e' term can never be 0 but e^(-∞) is 0 , why is this possibility neglected ,especially when the range of the function is not mentioned(8 votes)
- −∞ is not a number, so you will never have that in the equation. You can approach it (as in the case of a limit) but never reach −∞. Thus, you won't have a critical point where x=±∞ because there is no such point (it is only approached, but never reached).(15 votes)

- At0:54Sal mentioned that 'c' is a critical number 'iff'
*f'(c) = 0*or*f'(c) undefined*.

But I did few exercises on Khan Academy that ignores points when f'(x) is undefined and states that answer as incorrect.

*For example: *

When I'm doing a quiz on Khan Academy I see this:

*h(x) = e^2x/(x-3)*and a question asking to find critical points.

If you take derivative of that you would get:*h'(x) = e^2x(2x - 7)/(x - 3)^2*

Now clearly*h'(x)*is*0_ when x = 3.5, but also _h'(x) is undefined*when*x = 3*

Yet if you present that both*x=3.5*and*x=3*are critical number quiz returning answer than 3 is an incorrect answer.

It's not just isolated quiz, I did a bunch of them and every time you try to present critical x when f'(x) is undefined it renders my answer incorrect. What am I missing?

P.S.: I got it a minute after I submitted the question. Apparently x is not a critical point in this case, because when x=3 not only it's derivative h'(3) is undefined but also main function h(3) is also undefined. So it is not considered as a critical point in this case.(11 votes)- That's a big problem! Sal should've said and wrote: "c is a critical number of f iff f'(c)=0 or f'(c) undefined, and f(c) defined" instead.(1 vote)

- How can I find critical values of a function when the zeros of the first derivative of that function are imaginary numbers? For example, the function is 7x^3 - 3x^2 + x - 15.(6 votes)
- There is no rule saying a function has to have ANY critical points. For example,

f(x)= 3x+7 has no critical points.

Your particular function has no REAL critical points. Most likely, that is what an introductory calculus course would be asking about, so you would most likely be expected to say it had no real critical points.

However, it is completely valid to have nonreal critical points. All you do is find the nonreal zeros of the first derivative as you would any other function. You then plug those nonreal x values into the original equation to find the y coordinate.

So, the critical points of your function would be stated as something like this:

There are no real critical points.

There are two nonreal critical points at:

x = (1/21) (3 -2i√3), y= (2/441) (-3285 -8i√3)

and

x = (1/21) (3 + 2i√3), y= (2/441) (-3285 + 8i√3)(10 votes)

- At2:50, shouldn't d/dx e^-2x^2 = -2e^-2x^2*(-4x)?(7 votes)
- I put it into wolfram alpha and got: Possible derivation:

d/dx(x/e^(2 x^2))

Rewrite the expression: x/e^(2 x^2) = x/e^(2 x^2):

= d/dx(x/e^(2 x^2))

Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = e^(-2 x^2) and v = x:

= x (d/dx(e^(-2 x^2)))+e^(-2 x^2) (d/dx(x))

Using the chain rule, d/dx(e^(-2 x^2)) = ( de^u)/( du) ( du)/( dx), where u = -2 x^2 and ( d)/( du)(e^u) = e^u:

= (d/dx(x))/e^(2 x^2)+(d/dx(-2 x^2))/e^(2 x^2) x

Factor out constants:

= (d/dx(x))/e^(2 x^2)+(-2 d/dx(x^2) x)/e^(2 x^2)

Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x:

= (d/dx(x))/e^(2 x^2)-(2 x 2 x)/e^(2 x^2)

Simplify the expression:

= -(4 x^2)/e^(2 x^2)+(d/dx(x))/e^(2 x^2)

The derivative of x is 1:

= -(4 x^2)/e^(2 x^2)+1/e^(2 x^2)

Simplify the expression:

Answer: |

| = e^(-2 x^2)-(4 x^2)/e^(2 x^2)(2 votes)

- at4:28why the e^-2x^2 cant be zero ?(6 votes)
- The exponential function (when considered a function of real numbers) is always positive. That is, for every real number
`u`

, we have`exp(u) > 0`

. Now let`u = -2x²`

, which is a real number whenever`x`

is a real number. It follows that`exp(u) = exp(-2x²) > 0`

. Hence never zero.

The fact that`exp(u) > 0`

for every real number`u`

requires a proof in itself, but I will not provide one here (unless you really want me to).(5 votes)

- If you wanted to find a critical POINT (not number) would you just take the 'x' value -the critical number, and plug it into the original equation to get a 'y' value?(3 votes)
- Yes, that is how you would find a critical point. You get the x from the first derivative and the y from the original function.(3 votes)

- Doesn't f(c) need to exist for x=c to be a critical point of the function? This example doesn't highlight this part of the critical point definition.(3 votes)
- Hi there,

When defining a critical point at x = c, c must be in the domain of f(x). So therefore, when you are determining where f'(c) = 0 or doesn't exist, you aren't included discontinuities as possible critical points. Here is an example.

f(x) = x^(2/3). The domain here is all real numbers.

f'(x) = 2/3 x^(-1/3) or 2/(3x^1/3) . (or 2 divided by 3 cube root x)

x = 0 is not in the domain of the derivative, but it is in the domain of the function. Therefore, x = 0 is considered a critical point of f(x).

Hope this helps(3 votes)

## 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.