Question
Find the critical numbers of the following functions(a) g(r) = r' 8r? (6) f(r)(c) f (0) = 2sec0 + tan 0, 0 < 0 < 2T .Find the absolute extrema of the function O the given closed interval, 21 (a) h(r) = 5 - r" (-3, 1] () f(r) = 2r" 61, [0, 3], (c) f() = [-2, 4 T+2"' (d) h(t) 1-1, 6}, (e) g(r) = Ir+4, (+7, 1}
Find the critical numbers of the following functions (a) g(r) = r' 8r? (6) f(r) (c) f (0) = 2sec0 + tan 0, 0 < 0 < 2T . Find the absolute extrema of the function O the given closed interval, 21 (a) h(r) = 5 - r" (-3, 1] () f(r) = 2r" 61, [0, 3], (c) f() = [-2, 4 T+2"' (d) h(t) 1-1, 6}, (e) g(r) = Ir+4, (+7, 1}
Answers
Identify the critical points and find the maximum value and minimum value on the given interval. $$r(\theta)=\sin \theta ; I=\left[-\frac{\pi}{4}, \frac{\pi}{6}\right]$$
So let's start by taking a conservative. So we get to close Sign of data minus side of data, and I'll set this equal to zero to get to Cole Center. Feta. You could see China later. Dividing my coastline gets using detentions. They're so they don't Most people Teoh tangents, inverse of To, which is 1.107148718 Okay, so let's check or critical points. So that's tensions in verse of To as well as our endpoints. Okay, so in this case, we get one for our output. And then what about negative to pine? That's one as well. And then to sign of Tencent, inverse of to plus co sign of tension in verse of to So that gives me 2.24 So I see that this is our absolute max and then we have to absolute men values
For this problem in part A We're asked to find the critical points of the function F of theta equals to sine Theta plus coast data on the closed interval from negative to pi to two pi. So to begin, we want to find the derivative of or of F of theta there. So derivative of two. Cynthia is going to go to to coast data and derivative of coast that is going to go to negative sine theta. We want this to equal zero. So in turn that means that we want to our to coast data to equal sign data. Or alternatively, we need to to equal sine Theta over coast data. Which is equivalent to saying that we want to to equal tan theta, which in turn means that data is going to be tan inverse of two, which comes out to be first of all. That is going to be multi valued. So one moment here. So we'll have within our interval There will be 10 inverse of two, 10 in verse of two plus two pi. And then also we'll have 10 inverse of two. 10 Universe of two. Mhm minus pi. Or actually excuse me minus pi. Then Let's see we would also have 10 inverse of two plus pi. So we'll have four different points Where we'll find that at 10 inverse of two. We'll get a maximum and then we'll have We go forward by two pi. That will actually be outside of our bounds, But actually -2 pi is the one that we want there The one with -2 pi then that will give us a minimum. Or actually let me correct myself that's going to be a maximum as well. And then at tan inverse of pie and tan inverse of minus or tan inverse of two minus point, an inverse of two plus pi. We'll get minima where we'll find that F of tan inverse of two as well as at 10 inverse of -2. It will have a actually, I'll be careful here, a 10 inverse of two. It has a value of about 2.236 At 10 in verse of 2 -2 pi Has a value of equal or it has the exact same value 2.236. Then at F of tan inverse two minus pi Has a value of -2.236. And same deal for f of tan inverse of two plus pi. Okay.
Yeah. H. Of our equals one over R. And the interval is negative one 23 Well, each prime of our would be negative one over R squared. Uh huh. But wait a minute one over R. If we were to graph one over R it would look like this and like this. So the maximum value it goes all the way up to infinity and the minimum value it goes all the way down to negative infinity. So the max would be infinity, the men would be negative infinity. Or you could say there is no maximum or minimum. Thank you for watching.
Okay, so we want to find our fallen critical points. So let's start by taking a generative so we'll be using our quotient role. So this is three times T squared plus three minus two tee times three t sort out six teeth. Word over T squared, plus one squared. So what does that give me? That negative three t squared, plus three over the falling. Okay, so if we set this equal to zero when we feel that our critical points are at, um, we contract out of three so we just get negative t squared, plus one. So we get to use equal to plus minus one. Okay. And use give us this denominator. It gives us imaginary numbers so we won't take into account that, but we'll take into account our endpoints, and then let's find our following effort T. So I'll plug this into my calculator. Okay, so I get negative 1.2. No, I think one. We get negative 1.5 at one. I get 1.5, and at two, I get 1.2. So we see that we have an absolute men when t is equal to negative one and an absolute max when t is equal to one