5

Problem 9_ [1Opts ] For the function f below , YOIL IaLY present YOur respOnses to thc following questions using chart formats as covered in the COUISC ad using mn...

Question

Problem 9_ [1Opts ] For the function f below , YOIL IaLY present YOur respOnses to thc following questions using chart formats as covered in the COUISC ad using mnethods fromn this Math 71 class: f(1) 312 + 8r | 4Find Lhe intervals whcre the function is increasing/decreasing?Find all relalive mLXims? relativc: Ininims?

Problem 9_ [1Opts ] For the function f below , YOIL IaLY present YOur respOnses to thc following questions using chart formats as covered in the COUISC ad using mnethods fromn this Math 71 class: f(1) 312 + 8r | 4 Find Lhe intervals whcre the function is increasing/decreasing? Find all relalive mLXims? relativc: Ininims?



Answers

In Problems 65-72, verify each function has a zero in the indicated interval. Then use the Intermediate Value Theorem to approximate the zero correct to three decimal places by repeatedly subdividing the interval containing the zero into 10 sub intervals. $$ f(x)=x^{3}-4 x+2 ; \text { interval: }(1,2) $$

Okay, so we're going to use the intermediate value theorem to show That there is a zero in this function on the interval from 1-2. So the intermediate value theorem says that if we have a value on the interval, say the intervals from A to B. If we have a value of F F. A, that is greater than C, where C is just a constant, so it's equal to some value. And then we have at F f B, we have that F B is less than C. Then as long as this is a continuous, smooth curved function, which this is, then we're guaranteed to have a value at some um value in between A and B. I'll just call it F F. D where D. Is, and I'll even write it down where A is less than D, which is less than B. We're guaranteed to have a value here, that's equal to see. So really, what we need to show is that there is um that F of one is greater than zero or less than zero, since this works either way, and then F of two is the opposite. So if F of one is greater than zero and ff two needs to be less than zero, and ff of one is less than zero than ff two needs to be greater than zero. Then we can guarantee that there is a zero in between that interval. So let's look at F of one, This is equal to four times one of the third minus two times one squared minus seven, which is equal to four minus two minus seven, which is equal to um 2 -7, which is equal to negative five. So now we have that f of one is less than zero, so now we need to to be greater than zero to guarantee that we have a zero in that interval. So it's like that F F two, this is equal to four times two to the third minus two Times two Squared -7. So to to third is eight and if we multiply eight x 4 we get 32 and then two squared is four minus two is eight, and then we have minus 7, 30 to minus eight is 24. And then if we minus seven we get Um 17, so this is equal to 17, so therefore FF two is greater than zero. So if we use the intermediate value theorem, that means that on the interval, so on the interval Um from 1 to 2 we are here guaranteed. Yeah, to have FFC equal zero at some see value. Um in this interval, so as long as C is between one and two, since f of one was less than zero and ff two is greater than zero were um guaranteed to have some value or to have f f c where C is some value equal zero in between that interval.

Well everyone. So today we're going to observe ffx and answer for things about it. We're going to find where what intervals It increases and decreases on its local maximum men's it's intervals of con cavity and likewise it's inflection points. So let's get started here. So answer your first question where increases in decreases. The best thing you can do is observe the first derivative of prime affects. Because we know from the increasing decreasing test that if the first derivative is positive on some interval then the main function is increasing there and if F prime of X is negative on an interval then the main function is decreasing there. So let's take the throat. It really quick should be four X cubed minus four X. Simple. So um now what do we do with this? I wager that we should saw for one F prime of X zero. Because if we if we want to know where F prime of X is positive and negative, we should find the zeros of that function and then we should decide what the behavior of F prime is around that. I think that's the best way to do it. So let's say zero equals four, executed my forex. In fact, er for X out of each term, divide out our factor for and right here there's a difference of squares. So we know this is x times X plus one times X minus one. Now I have are three zeros, so X is gonna equal zero -1 and one awesome. So to observe the function around the zeros, let's draw a quick number line, these are excess. Let's pick some exes around our zeros to observe what's going on. So let's pick some like easy the whole list numbers we could possibly picks. Let's do like negative too. On the deep end here and two over here I might do negative 1/2 and one half because it's not much easier between zero and 1. Right, So, great. First of all, solve for what? Why what why value F. Prime has when X is negative two. Do that really quick. Should Equal -24. Let's do as prime of negative one half. Next Should be 3/2. Let's do our next Thing here. one half Should be -3/2 and last but not least. Let's do F prime of two. And this should be positive 24. So let's see what's happening here. We've got this number is what this number is negative, this number is positive, this number is negative and this number is positive. So We know that around here before x equals negative one. We know that f prime here equals -24. So we know that F prime is negative here. So we know f is decreasing here up from negative infinity to x equals negative one. We know that here. Uh F prime was evaluated to be a positive number, right? Because we've got three houses are answers. So F should increase here because of prime is positive. Next we've got Prime was -3 House here. So after prime was negative. So F decreases here Anyways. But at least around f prime of two, F prime was positive. So if increases here and that should tell us all, we need to know about where our function F increases and decreases. So let's see, should increase from -1-0 and from want to infinity and should decrease From negative infinity to negative one And from 0 to 1. Great. So that's our first item checked off awesome. Now for local maxes and mints, we basically already have everything going for us here, right? Because the local maximum in happens where the function changes from increasing to decreasing and a local men would happen when the function is changing from decreasing to increasing. Right? And we already have those numbers going on here. So there you go. Let's evaluate if a negative one off of zero and a point. Right? So let's see really quick. We've got F one equals should be too F of zero. Should be. and of course this is three and f of one should evaluate to, I forgot to write the negatives and the -1. Oops well these are both to uh to end power. So they do end up evaluating As we can see two. Also why equals two. So we have one local max at see at 03 and we have to local men's at -1, 2 and 12, awesome. That's number two done. So now we need our intervals on the caddy and inflection points. I think I'm going to start this point on the next page. So we have more room. All right. So for intervals of can cavity, we're going to be looking at the second derivative for the con cavity test. And when you do come cavity tests, when the second derivative is positive, then our main function is khan gave up and when the second derivative is negative then our functions con game down. So let's do a quick second derivative here. Just take the derivative of F prime. So that's going to be 12 X squared minus four. And if we set this to zero to find where are zeros are, we'll see zero equals three, X squared minus one, 1/3 equals x squared and X is gonna equal plus or minus one over root three. A lot of math textbooks at about this level are pretty afraid of writing radicals in the denominator for some reason. So, I'm going to write this route through over three. Just a teachable moment to show you why because this ends up happening a lot in trig and like my brain couldn't put it together why people were doing this. But it's really simple. One of the route three times route through over Route three Equals 3/3. Right? Like it makes sense. I just needed someone to show that to me. I was like, why are you making this substitution? You know? But I don't know. Very simple algebra. People are just afraid to put radicals in the denominator for some reason. I don't know general math moment. Right there for you. Cool. So let's do our number line again. Sorry about the cut. I realized these were missing for some reason. Okay, so um All right. So we're at negative route 3/3 and reserved three And let's pick some easy. X is really quick. Let's do zero -1 & one. Okay, so let's plug in for these. Really quick to see what happens with F double prime. So F double problem negative one is equal to eight After the prime of zero Is going to be -4. Can you put a zero in here? And F double prime of one is going to have the same signage as The -1 here because -1 Squared. So this is also going to be eight and you can see that um this one is positive, this one is negative and this one is positive. So we can see that change and sign an F double prime happening again. So, all the way on the left, from negative infinity to negative 3/3, we've got positive f double prime. So F is concave up, F double prime is negative in here. So F is concave down and F double prime is going to be positive around here. So F is going to be concave up, awesome. So let's just really clearly right, our interval is going on. We are con cave up. Uh huh. From negative insanity to negative roots 3/3, And from route 3/3 to infinity. And we're concave down From negative route 3/3 to root. Seriously? Yeah, awesome. So the last thing we need is our inflection points and we have that really clearly written here because we already know where F. Is changing con cavity, right? It's on our zeros that we already found because we know that they change from positive to negative across each one. So all we gotta do is just plug Uh negative route for over three and root for over three into F. So let's do that really quick F of negative route 3/3 is equal to 22/9 And f. of 3/3. It's going to be the same thing, right? Because these powers are to end, so the Negativity is canceled. So it's going to evaluate to the same thing 22/9. So we have to inflection points at negative 3/3, 22 or nine. And routes through three 22/9. And that's it. That's the end of our problems. Ah Thanks so much for watching. And I hope it helps.

We have a whole sequence of problems. Um that are so some questions about this uh this graph here that I've tried to reproduce from the book, um it's obviously piecewise continuous. It's not smooth because it looks like we have a corner there. So linear here, linear here, but with a constant Y. And then some kind of probably more like a, like a cubic or cortical function. Um Over here. They don't give it, they don't tell us what these are. We could yeah, I think we have enough points. Let's see here, we have 12345 points. We could figure out a quartet polynomial. We probably want to find this slope here. So we could figure out a Quinton polynomial on that word. Have a zero slope here and go through all these points if we wanted to. But that's not what we were asked. So we're just asked a bunch of questions about this. So I'm just gonna go through all of them in this one video, because they're all very much related and there are, you know, I have to do with this this graph here. So the first question is fine. F zero and effort negative six. Well F zero, that's when X equals zero. So we come up here and we're at my at three and then at negative six we're out here and so they've labeled this point here, X is negative six, then why is minus three now they ask us is um is F of three positive or negative? So let's see here F of three is somewhere in this in this region here and so it's positive and it's probably, you know, just again, they didn't label the point but it looks like it's three for this whole region between zero and four. An f of minus minus four, so minus three were at zero minus five, we're at minus two. So if this is indeed a line then minus four were at minus one and then clearly f of my f one of X equals minus four is clearly negative. It's on it's to the left of here. No, they ask us, let's see here. Um for what values of X is F of X zero? Well f of X is why here? So we have zero here, zero here and zero here. So there's three points. And they are when X is minus three, when X is six and when X is 10 to those three points, then they ask us for what values of X. F greater than zero. Well we can see here that it's greater than zero here and also here. And I should probably say that this should probably change this to just assume that this doesn't get extended. So we have it's greater than equal to 10. Less than equal to 11. So from here to here, right, we have X equals minus 3 to 6. So and they said greater than all Right. So I should have I should just have greater than it's not equal to um And then from here to here, you know, it goes from 10 to 11 over this region. Little region here, we also have positive values. Then I asked for the what is the domain? Well, assuming that, you know, this doesn't continue on in any way that this is just, you know, it's only defined over this region. The domain goes from X equals minus 62 X equals 11. And they asked us for the range and the range is the span of y values. So it looks like the smallest Y value we have is minus three and the largest value we have is three. So that's the range. Then we'll ask for the X intercepts. The X intercepts are actually just the points where y equals zero. So we have the exact same points here minus three at six and 10. They ask us for what is the wiring rcep? Well that's when X zero and we see that in X zero. They tell us that why is three? Yeah. And how often does the line why it was gonna have intersected graph? Well, Y equals one half. If we go down um you can see that it's gonna be down something like this. So it looks like about three. And so we know it intersects somewhere here as it this crosses it, we know what intersects somewhere here as that crosses it. And over here at this point this goes from um Y equals zero to Y equals one. So one half is right in between there. So we know what crosses it there too. So that crosses three points. How often does the line cross? Um The graph crossed the line X equals five and I thought why it was five, so this is wrong. And so let's do that. Y equals five. Um No X X equals five. So it just crosses at once. And in fact if it's a function, if it's a function then it can only cross any any constant value of X at one at one point. Otherwise it's not by definition of function. So yeah, process somewhere here if this, well I don't know what kind of curve this is. So I would speculate on what the Y value is there then how for what values of X does F equal three? Well, it looks like again assuming this is all horizontal here, it looks like it's F equals three for this entire span here from 0 to 4. Mhm. Than for what values does F equal minus two? Well we have we have a point over here minus five minus two, so we're given that one and we're told this goes through eight minus two. And again this just this double cross comes back down here. So if we look at um for why it goes minus two, we just touched this point here and I'm assuming that's a minimum it appears to be, so it doesn't go beyond it somehow. Um you know, doesn't go below and then over here, so those are the two points X equals minus five and eight. And then for and what interval is the function increasing? What intervals? So obviously it's increasing here. Right. And it's increasing here. So they have yeah. Um And so this region here is minus 6 to 0. That's not increasing here. Nor is it decreasing? We'll come back to that later. And it looks like from here from 8 to 11. It's also increasing. Did they ask about when there's a decreasing? Well it's decreasing from here to here and that goes from 4 to 8. And they asked when is a constant? Well that is it's constant between zero and four. And then they asked, when is it non increasing? Okay, so that means constant or decreasing. So basically have the union of this set and this set which is 0 to 8. So from here, other way to here is non increasing. And then they ask if it when is it non decreasing? Well, again, that's the intersection of the constant region and this region here, so that winds up being minus 6 to 4. So here to here, it's non decreasing. And then over here, 11 or 8 to 11, it's also increasing or non decreasing. So those are all the questions they ask us about this this one graph. Um, so hopefully there should be pretty simple for you to do at this point, and hopefully none of this was kind of a surprise, um, to looking at this chart here and in asking all these questions.

With a given problem, we want to consider the function F. Of X. Equalling X to the 4th minus X cubed. Uh huh. Class acts right next to. We're looking over the interval from 1-2. So f of one, I'm gonna give us negative. One effort to is going to give us eight. Um and then we're going to bring this closer, so 1.3. Um and then about 1.4. So actually that's very close because this is negative and this positive Gonna be closer though to 1.3. So if we bring this down to 1.31 At the me here, so 1.305, It's also negative. So we would say 1.306307. Um so about 1.308 or 1.309 Based on this 1.309 will be a better approximation.


Similar Solved Questions

5 answers
Draw the structure of the following: (25 pts) chloromethyl methyl ether acetophenone hexanedioic acid any lactone triphenylphosphine
Draw the structure of the following: (25 pts) chloromethyl methyl ether acetophenone hexanedioic acid any lactone triphenylphosphine...
5 answers
Find an equation in and ! for the line tangent to the parametric curve x(t) = 8e"" y(t) = (t_ 9)? at the point (<.y) = (8.81}
Find an equation in and ! for the line tangent to the parametric curve x(t) = 8e"" y(t) = (t_ 9)? at the point (<.y) = (8.81}...
5 answers
Consider the following_x(ax _ 1)4 Let u 4x3 Find du.dudxIndicate how the limits of integration should be adjusted in order to perform the integration with respect to u. (Enter your answer using interval notation:)[0, 1]Evaluate the definite integral:
Consider the following_ x(ax _ 1)4 Let u 4x3 Find du. du dx Indicate how the limits of integration should be adjusted in order to perform the integration with respect to u. (Enter your answer using interval notation:) [0, 1] Evaluate the definite integral:...
5 answers
A hair salon reports that on seven randomly selected weekdays; the number of customers who visited the salon were 15,20,19,12, 36,16 and 50. This is our population and we can assume normality in this case.Construct a 90% confidence interval for the average number of customers who visit the salon on weekdays 2. Interpret the results
A hair salon reports that on seven randomly selected weekdays; the number of customers who visited the salon were 15,20,19,12, 36,16 and 50. This is our population and we can assume normality in this case. Construct a 90% confidence interval for the average number of customers who visit the salon on...
5 answers
Points A and B are connected to two terminals of battery with electromotive force €R1-2 Ohms;R2 = 4 Ohms;R3 = 6 Ohms;R4 = 8 Ohms_E = 12 V: Find the voltage drop on resistor R4_5.538 V0.412 V1.954V
Points A and B are connected to two terminals of battery with electromotive force € R1-2 Ohms; R2 = 4 Ohms; R3 = 6 Ohms; R4 = 8 Ohms_ E = 12 V: Find the voltage drop on resistor R4_ 5.538 V 0.412 V 1.954V...
5 answers
Refer to the vectors given below "-[Hk--[Hk-[#-[k-[H-[#-[#-[:J Compute projs Uz where S span{u3' u4}-projs U2
Refer to the vectors given below "-[Hk--[Hk-[#-[k-[H-[#-[#-[:J Compute projs Uz where S span{u3' u4}- projs U2...
5 answers
Does the following sequence converge or diverge? Ifit converges_ what rumber does converge t0?On = (-1)"ConvergesConvergesDivergesConverces 23Converges
Does the following sequence converge or diverge? Ifit converges_ what rumber does converge t0? On = (-1)" Converges Converges Diverges Converces 23 Converges...
5 answers
Explain why solutions of HBr in benzene (a nonpolar solvent) are nonconductive, while solutions in water (a polar solvent) are conductive.
Explain why solutions of HBr in benzene (a nonpolar solvent) are nonconductive, while solutions in water (a polar solvent) are conductive....
5 answers
Chapter 11 Section 11.6 Go Tutoria Problem 011Find the equaticn ofche lane containing the pcints (-7,1,1), (0.2,4),and (1,0,- 2)Edit
Chapter 11 Section 11.6 Go Tutoria Problem 011 Find the equaticn ofche lane containing the pcints (-7,1,1), (0.2,4),and (1,0,- 2) Edit...
5 answers
Write each ratio as a fraction in simplest form. See Example 3 .56 yards to 64 yards
Write each ratio as a fraction in simplest form. See Example 3 . 56 yards to 64 yards...
1 answers
Solve the linear programming problem.MaximizeP=30x+40ySubject to2x+y≤18x+y≤10x+2y≤16x,y≥0a) What is the maximum value of P?b) What are the coordinates of the corner point wherethe maximum value of P occurs?
Solve the linear programming problem. Maximize P=30x+40y Subject to 2x+y≤18 x+y≤10 x+2y≤16 x,y≥0 a) What is the maximum value of P? b) What are the coordinates of the corner point where the maximum value of P occurs?...
5 answers
Verify that the hypotheses of the Mean–Value Theorem aresatisfied for the function f(x) = x^3+x−4 on the interval [−1, 2] ,and find all values of c in the given interval that satisfy theconclusion of the theorem.
Verify that the hypotheses of the Mean–Value Theorem are satisfied for the function f(x) = x^3+x−4 on the interval [−1, 2] , and find all values of c in the given interval that satisfy the conclusion of the theorem....
5 answers
Use Jacobi iterative method, wi Use Jacobi iterative method, with initial guess x(O) ,y(O),z(O) = (1,2,2), to solve the system ~2x +y + 52 = 15 4x - 8y + 2 =-21 4x -y+2 =7
Use Jacobi iterative method, wi Use Jacobi iterative method, with initial guess x(O) ,y(O),z(O) = (1,2,2), to solve the system ~2x +y + 52 = 15 4x - 8y + 2 =-21 4x -y+2 =7...
5 answers
Com'asse5smtents Je Pobart Mnomis Lakevillo HomeUSHistory SlateSubmit TestReader ToolsUMjoSave ExtLevel Farces and SolutionsUypeconec Amaueacm box Expressrnsamsignificant figures_You lorm Water vapa moing axygen and hydrogen 730 C in Oz(g) ZHz(g) 2H2O(g)4-ter conanet This the equation Ior the reactionparnal 0kes5uteoxyron belore the reachum122 kilopascalsthete exco5' hydrogen. How many moles ol wateronmed?he (nactn prod ucesmole 5WaterResetNext
com'asse5smtents Je Pobart Mnomis Lakevillo Home USHistory Slate Submit Test Reader Tools UMjo Save Ext Level Farces and Solutions Uype conec Amau eacm box Express rnsam significant figures_ You lorm Water vapa moing axygen and hydrogen 730 C in Oz(g) ZHz(g) 2H2O(g) 4-ter conanet This the equat...
1 answers
B) For the function f (1) =x-x.0 < x < 1, obtain the Bessel series of the form f (x) = a1J,(_*)+azJ,(2*) where 1n 's are the positive Zeros of J,(x)
b) For the function f (1) =x-x.0 < x < 1, obtain the Bessel series of the form f (x) = a1J,(_*)+azJ,(2*) where 1n 's are the positive Zeros of J,(x)...
5 answers
Mark True or False_(Negative marking may be used)The X-axis is tangent to the polar curve r = cos(e /2) at 0 = 3TChoose ..Ifr= then the conic section is a parabola: 3+ cos(0_Choose ..The arc-length of the polar curve r = 2 V0 for 0 < 8 < T/2 is given by T/2 L-2f" V1+- 40 d8 .Choose:
Mark True or False_ (Negative marking may be used) The X-axis is tangent to the polar curve r = cos(e /2) at 0 = 3T Choose .. Ifr= then the conic section is a parabola: 3+ cos(0_ Choose .. The arc-length of the polar curve r = 2 V0 for 0 < 8 < T/2 is given by T/2 L-2f" V1+- 40 d8 . Choose...

-- 0.021631--