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Problem 45 Easy Difficulty

A formula for the derivative of a function $f$ is given. How many critical numbers does $f$ have?

$ f'(x) = 5e^{-0.1 |x|} \sin x - 1 $

Answer

The graph of $f^{\prime}(x)=5 e^{-0.1|x|} \sin x-1$ has 10 zeros and exists
everywhere, so $f$ has 10 critical numbers.

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Video Transcript

Okay, let's go ahead and answer this question. So the question tells you why close to five e to the native 50.1 absolutely affects time. Sign of X minus one. Um, how many critical points do we have when this Why is actually a derivative of a function f Okay, so to find a critical point, what's important to remember is that, first of all, it has to be a value of zero. And second of all, it has to actually be a sign change. So F prime has to be zero in the points around. It has to be from negative to positive or positive to negative. If you take a look at the graph, you can see that it's not one of those, um, shapes where it's on Lee touching the X axis. So it's very likely that it's not going to be a point where if prime is equal to zero yet, it's not a critical point. So we're not gonna have to worry about those things now. It's very important to see that there is a negative one right there, because if it's not, you can clearly see that, um, it zero is the average value off this function, and it'll very likely have infinitely many points as a critical. However, the negative one shifts it downwards. So we have to be very careful with how many zeros this thing has. Okay, because let's do something really interesting right now. Five need to the Maybe there's 0.1. Absolute value of X shifted one ground, as you can see. Oh, let me try to change it. There you go. This is the number that is multiplied in front of the sine X. So it behaves like it amplitude. So how much up and down it goes? Depends on the value of this guy, Okay? And it basically traces on top of the highest point where this Greiff graph can go. And as you can see, the further you go towards the right side, it gets closer and closer to zero. So it seems like at a certain X input, the height is not going to be tall enough that it can touch the X axis even though the graph wants to get close to it anymore. Okay, That's one of the interesting things that you could see. Now we're going to count how many points it actually crosses through. Um, why equals to zero. So we're going to start here. You gotta be very, very careful here. If you go far enough, it looks like it actually touches the X axis. But if you get closer, you can see that it actually is not. So this point, even though it's close, it actually doesn't count as a, um zero. So let's count. 12 34 56 789 10 and Tenet ISS So you can see that all of the's 10 points are ex intercepts, and it goes from negative positive to negative here, negative to positive there. So it's a sign change. So, indeed, all of these zeros represents a critical point off F because F Prime is equal to zero at those points. So to answer the question, how many critical critical points should f have? Well, it's going to be 10, and that's I answer this question