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1/2 points Previous AnswersConsider the function F(8) 2sin28 2cose(a} Find f '(8)4cos (8) + 2)sin (0)(b) Find the critical point(s) of f(8) on the interval 0 &...

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1/2 points Previous AnswersConsider the function F(8) 2sin28 2cose(a} Find f '(8)4cos (8) + 2)sin (0)(b) Find the critical point(s) of f(8) on the interval 0 < 0 < T. If necessary, enter your answers as comma separated lists3,10/1 points Previous AnswersIf U and V are positive constants,; find the critical point of the function_ Use upper case letters, F(t) = Uet Ve-~t(log (u) log(~15 pointsUse f(x) 30x2e 0,.3x to answer the questions below.(a) Find 'x)(b) Find the exact critica

1/2 points Previous Answers Consider the function F(8) 2sin28 2cose (a} Find f '(8) 4cos (8) + 2)sin (0) (b) Find the critical point(s) of f(8) on the interval 0 < 0 < T. If necessary, enter your answers as comma separated lists 3,1 0/1 points Previous Answers If U and V are positive constants,; find the critical point of the function_ Use upper case letters, F(t) = Uet Ve-~t (log (u) log( ~15 points Use f(x) 30x2e 0,.3x to answer the questions below. (a) Find 'x) (b) Find the exact critica points of and classify each as local minimum, local maximum, neither: (smaller value} This critical point is Select- (larger value) This critical point is Select--



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a. Find the critical points of the following functions on the given interval. b. Use a graphing device to determine whether the critical points correspond to local maxima, local minima, or neither. c. Find the absolute maximum and minimum values on the given interval when they exist. $$f(\theta)=2 \sin \theta+\cos \theta ;[-2 \pi, 2 \pi]$$

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.

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

Okay, so let's start by finding are critical blinks so f prime of X. That's equal. Teoh leaving them that green. And then we have minus 2 1/2 I know we'll set the sequel to their all. Um, and now it's odd. 1/2 on both sides on multiply both sides by hey, over three. Do we have that? X squared is equal to a vital about two. That's for over three. And now let's take the square root herbal side. So we have X equal to plus minus two over square root. Nobody. Okay, what is shoes? Square root of three. 1st 2 divided by square root of three. Um, this is approximately plus minus one points. 1547 So instead, our domain is from negative. 123 So we're not including this negative point here, since that's negative 1.1, which is large, it and negative one. Okay, so it's we only have one critical point. That is when X is equal to positive. Two over square root of three. Now looking. Yeah, this on a graphing calculator, we see that we have a local men at that point

For this problem we are asked to find the critical points of the function F f T equals three T over T squared plus one on the closed interval from negative to to to and for so actually let's just start there. The first thing that we want to do is take the derivative of a function function with respect to T. So we can do this either as product or quotient. I'm going to do this essentially treating it as three tee times one over t squared plus one. So we'll have derivative of three tee times T one over t squared plus one. So we have a three over t squared plus one and then we have plus. Now we need to apply chain rule so we'll have plus two times three T. So that will give us a 60 cubed divided by Or actually that will give us a negative 60 cube divided by t squared plus one To the power of two. Now putting everything together, we can simplify this down to negative three times t squared um plus one. One moment here. Actually yeah, we can we can simplify this down to, you need to be careful about that negative sign. It should just be three times T square, never mind negative three times T squared minus one divided by t squared plus one. All squared. If we want to solve this for when it equals zero then we just really need the numerator to equal zero which means that we need T to equal plus or -1. Now when T equals plus one, 12, right? Just X equals one. We get a local or actually that is the absolute maximum. And when X equals negative one, we get the absolute minimum. Where the values are f of one equals 3/2, And f of negative one equals negative 3/2.


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