If you're seeing this message, it means we're having trouble loading external resources on our website.

Εάν είστε πίσω από ένα web φίλτρο, παρακαλούμε να βεβαιωθείτε ότι οι τομείς *. kastatic.org και *. kasandbox.org δεν είναι αποκλεισμένοι.

Κύριο περιεχόμενο

Finding critical points

Sal finds the critical points of f(x)=xe^(-2x²). Δημιουργήθηκε από τον Σαλ Καν.

Θέλετε να συμμετάσχετε σε μια συζήτηση;

Δεν υπάρχουν αναρτήσεις ακόμα.
Μπορείς να διαβάσεις στα Αγγλικά; Κάνε κλικ εδώ για να δείτε περισσότερες συζητήσεις που συμβαίνουν στην αγγλική ιστοσελίδα της Khan Academy.

Απομαγνητοφώνηση βίντεο

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.