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Problem 14

Answer the following questions about the function…

Problem 13

Answer the following questions about the functions whose derivatives
are given.
a. What are the critical points of $f ?$
b. On what open intervals is $f$ increasing or decreasing?
c. At what points, if any, does $f$ assume local maximum and minimum values?
f^{\prime}(x)=(\sin x-1)(2 \cos x+1), 0 \leq x \leq 2 \pi


x=\frac{\pi}{2}, \frac{2 \pi}{3}, \frac{4 \pi}{3}



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Video Transcript

So here we're looking at a problem that is the first derivative of a function, and we need to find its critical points where it's increasing, decreasing and its local min and max values. So if we analyze our first derivative, which we have right here, we need to find out where it is equal to zero. Um, so we would have values equal to zero when the inside of these parentheses are respectively zero on their own. Um so if we look at our sign of X minus one, that's going to be equals 20 when side of X equal to one on our side of access, only going to be equal to one at pi over two. So let's start writing down our critical points here. Get prior to so it's going to be one of them. Now let's look at our toe to co sign of X plus one, so it's going to be equal. Teoh zero. When co sign of X equals negative 1/2 because negative 1/2 times two is negative. One a negative one plus one is zero. So coastline of X equals negative 1/2 um, at four, pira three and flight 55 or three s. Let's write this down for Pirate Three and five pi over three on. Now we need to find out where this function is going to be increasing or decreasing. So let's start out by putting in, um zero. For example, if we put in a zero Ah, our sign of X is going to be equal to zero minus one. So it's gonna be a negative number and there to co sign of X is going to be equal to two plus one. It's going to be positive number. So that would mean that it's going to be decreasing from zero to our pi over two. It s all right. That right here. So it's decreasing from zero to pi or to announce. Look, I in between Pirate two and four pi over three eso In between those numbers, we could just plug in, um, pie. So let's plug in pi to our sign. That's going to give us a zero minus one. That's gonna give us a negative number. And then if we plug in pie two or co sign, that's gonna give us a negative, too. So it's also going to give us a negative number, meaning that it's increasing from pi over 2 to 4 pi over three. All right, so now we need to pick a number that is in between our pyre or two. And I mean are four pi over three or 55 or three slits to three prior to eso. If we put in a three pirate to tow our sign of X, we get negative one minus one. So I give us a negative number and then if we plug in three prior to two hour to coastline, we get a zero. So give us a positive number. Meaning that dysfunction is decreasing from four pi over 3 to 5 pi over three. So let's write that down. He got for pi over from 3 to 5 pi over three. And lastly, we need to find out what this function is doing in between seven pi over four. And I mean sorry. Five pi over three and two pi. So you do that by let's plug in on seven wire before so we would end up getting, um, route to over two, which is equal to 0.7 slight number is going to be less than one. So if we, um, plug in our route to over two to this, we get our negative 20.7 minus one, so it's gonna give us a negative number. And if we Blufgan, um seven. By report of this, we get a two wrote to uh and then the twos would cancel out. You get a positive two, and then that number would end up being positive. There should have a negative, and a positive sore function is going to continue to decrease from 55 or 3 to 2 pi. And let's write that out right here. Oh, I'm sorry I missed that. I should be a three and then to to buy. All right, So now let's look at the local min or Max values, So we would only have that when it's switching signs. Um, so when ah function is going from decreasing to increasing, it's going to look like that and they're gonna get a local minimum going from increasing to decreasing. It looks like that and you get a local maximum. Eso for function is going from 0 to 2 pi on its decreasing, and then it starts increasing. Um, after our pirate too. So you gotta think decreasing to increasing thing. So at, uh, X equals higher or two, we'd end up getting a decrease in the increasing a local men. And then if we continue to look, our function is increasing from pi over 2 to 4 pi or three and then it starts decreasing afterwards. So we get something looking more like this and that our four pi X equals four pi over three, we would have a local maximum, and that is thelancet part of the problem.

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