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The local (relative) extremum (maximum, minimum) of flx) =xt 8x3 $ 1is (are)Select one:f(0) = 1 local maximum and f(6) = -431 local minimumf(6) = -431 only; which i...

Question

The local (relative) extremum (maximum, minimum) of flx) =xt 8x3 $ 1is (are)Select one:f(0) = 1 local maximum and f(6) = -431 local minimumf(6) = -431 only; which is local minimumf(o) = 1 only; which is local maximumNone of these{(0) = 1 local minimum and f(6) = -431 local maximum

The local (relative) extremum (maximum, minimum) of flx) =xt 8x3 $ 1is (are) Select one: f(0) = 1 local maximum and f(6) = -431 local minimum f(6) = -431 only; which is local minimum f(o) = 1 only; which is local maximum None of these {(0) = 1 local minimum and f(6) = -431 local maximum



Answers

Find all values of $x$ that make the function $f(a)$ a local minimum and $(b)$ a local maximum. $$f^{\prime}(x)=x^{3}(1-x)^{2}$$

Mhm. This time we're given the derivative. And so we need to start by just setting it equal to zero. So we set that equal to zero and we get X. Equals zero. X. Equals one. Now we need to take the derivative second derivative to see if these are maximum or minimum. So driven to the first times the second plus the derivative of the second times the first. Just simplifying. Although I really don't need to simplify. Let's just do the second derivative of 03 times zero. That's zero, second one's time zero. Also the second derivative of zero zero E. Second derivative of one also is going to be zero also zero. Mhm. Yeah two. So I don't necessarily know whether these are maxima or minima or possibly points of inflection. So I uh did something a little more advanced. I actually did the integral. Um And then plotted what it would have looked like. Um And so the actual function would look something like this. Um Yeah so at zero and one it actually is a minima. So these are both minima. Um We can't really tell based on the data uh what the vertical translation would be, so that graph could be up or down uh could be translated or shifted up or down. Um So those two are minimum.

All right, so they're giving us the derivative. Okay. And so we would have uh extreme A at X equals one. X equals two. X equals three and X equals four. Um So what I'm just gonna do is I'm gonna put in f prime of one, wait a minute. F prime of one would be zero. So let me not do that. All right. I think what I'm going to do because I want to do the second derivative is I'm just going to do foil method with the first two. It's going to be X squared minus three X plus two. And then I'm going to do foil with the last two, X squared minus seven X plus 12. Now we'll do the second derivative derivative of the first. I'm going to put negative in front of everything by the way, derivative of the first, which would be two x minus three. Yeah. Times the second plus the derivative of the second, times the first. Okay, so this is actually going to give me negative two x minus three. And then I'm going to separate this out again. X squared minus seven X plus 12. That's just x minus three. Times x minus four plus two X minus seven X minus one X minus two. Okay so we can put in f of one f of two of three of four F. Of one. I'm just gonna take a look at each of these terms. Um No I actually think I'm going to have to put in numbers for this. Mhm. Four Yeah. Okay negative two minus three is negative one. One minus three negative 21 minus four is negative three, two minus seven is negative five. Um One minus one is zero. So that all goes to zero. So negative times negative is positive times another negative is negative times another negative is positive so that gives you positive and so that would be a minimum. Okay? Now he's the same thing but I notice here at the at the end it's going to be plus zero again. So negative four minus three is one. Two minus three is negative. One to minus four is negative two. So this is going to be negative which would make it a maximum. Okay. Now I notice on this one. First term is going to be zero. This one too. First term will be zero. Second term two times three is 66 minus seven is negative one. Three minus one is two and three minus two is one. That's going to be positive. So it's gonna be a minimum. And last one, eight minus seven is 14 minus one is 34 minus two is two. That's going to be negative. Which is a maximum. All right. That's it. Thank you for watching. Okay.

Yeah. Alright, we're giving this function which is a derivative. And so uh if we set it equal to zero we get X equals one or two or three or four. So now I'm going to take the second derivative Mhm. I'm going to use the product rule from multiple terms. So I'm gonna take the derivative of the first which is to whoops two, X minus one. And then the derivative of the inside is just x minus one times a 2nd, 3rd and 4th times everything else plus derivative of the second, which would be to x minus two times everything else. Whoops. That X minus three was not squared. Okay. Plus derivative of the third term which is x minus three, Drew. That's just one times everything else. Yeah. Plus derivative of the last term which is just going to be one X minus three. Thinking about putting those last two ones together, X minus four times something. Plus x minus three time something. Well, let's not worry about that because I'm just going to do F double prime of one. Well, they all have an x minus one term in them, so that would be zero f double prime of two. Well, they all have an x minus two term in it, so that would be zero f double prime of three only. This does not have an x minus three in it, so that's going to give me three minus one is two, so two squared three minus two is one and three minus four is negative one. So negative. So that will give us negative four F double prime of four. Only the last one doesn't have an X minus four terms, so four minus one is three, Four minus two is ah for us to and four minus one. Okay, so that's going to give me nine times four, which is 36. Okay, clearly since this is positive, it would be a minimum, and since this is negative it would be a maximum. Um These two are inflection points. Thank you for watching.

All right, So we're being told to find a cubic function f of x equals X cubed, possibly x squared. Plus the 8. 50 that has a local max of three at X equals negative two in a local men of 0 64 1. So that's all the information we're given. So the first thing we want to do is take the derivative. So we want to take the derivative of primer back, and this will give us three a x squared plus b not B expert going to be to be xr to be x plus c. And so, um, the reason I took the derivative is because we're told some information about the derivative. Believe it or not, we're told that the local Max, um, and a local men So what is the local Max and local men means that there has to be a point at local maximum ensures that F prime is equal to zero. So we're told that at these values, negative two and one, So I have prime of negative. Coup is equal to f prime of one because that's when we have a local max. So when a local massacres, when it goes like this or in the local ministers like this. But both have a derivative zero. So we know that this fact is true. Mm. So what do we do now? We Well, we can just plug in our values into X for F prime. So we get three a, then you get negative. Two squared. No, no. Um, no, no, I'm sorry. This is going to be, uh So what? We're going? Yeah. So we're going to plug in. We're gonna plug these numbers into, uh, X right here. So we're gonna get three a and the negative two squared, plus, um, to be and then negative too. And then plus C and then we're gonna plug in one. So for one, we just get three a plus two b plus c. And then once you simplify this, you get 12 a minus four b equals three. A push to be because the sea is canceled. And now we can solve for either air B. I solve for B in this case. So he's all for B. You get three have. Hey, so now we what we do as we, um we know that f prime of one. So I'm gonna go ahead and put in red. So this is our second step. So we know that f prime of one is equal to zero. Yeah. And so what we're gonna do is we're gonna set, um three a X squared, just to be x plus C equals zero. So equal zero. And we're gonna plug in one. So this is going to come out to be, um, zero. It's equal to three. A plus to be. Let's see. Well, I'm gonna bring this on to the next page. I'm gonna rewrite this so it's gonna be zero. Is he called. I'm gonna go ahead and continue the red color to show are different. Step Oops, sorry. Zero is equal to three. Um, a plus to be plus c now, reminder. We saw for B so we can actually plug in or Newbie. So zero goes three a plus, two times, 3/2 a and m plus c. And now what we can do is we can just all for C galaxy is gonna give us negative six A. So now I'm gonna circle some of the important, um, constant. So we have b go to three half a and see you to go the negative six a. Now we're gonna move on to another step, and that is f of negative two. So what we're going to do so have a negative too. Now that we know the values of, um hey, we can actually go ahead and start plugging this into our original function. I mean, we know the value of B and C, so we're gonna go ahead and start plugging this into our original function. Yeah, so we're going to get f of negative two. So that's gonna give us negative tube cube times a and then plus negative two squared and in times be would be is three half a and then we're gonna get Plus, this is gonna be negative two times, see, and then plus D. And this is equal to three because this is told to us effort at X equals negative two. We get three mhm. We're going to hold on to this value right here and now we're going to solve for f of one so half of one. Well, that's going to be a plus three half a, which is our B. But now it's three half a rewritten then plus C and then plus D. And now this can be rewritten as five half a, um, plus the plus d and this is all equal to zero. Then we also know that f of negative two. You can also, um, So what we can also do now is we're going to do, um, f of negative two minus f of one. And that's gonna give us three minus zero, which is still equal to three. So f of two minus f uh, one. This is delicate three. And then this is equal to negative to a minus five. Half a managed to see minus c. So I did some substitution here. Um, Now we can get another value for See that we can rewrite to see negative one minus three, half A. We also know. So now we have to seize. Now that we have to seize, we cannot just solve for what number A is. Remember, we succeed equal to negative six A. So now we have two versions of C, so we can say negative. Six a is equal to negative one minus three half. Okay. And if you saw for this, you get a you go to to ninth. Wow. So Oh, that's great. Um, now we know a So now that we know A we also know that B is equal to three half times a so three half and then we know we just saw for a is 29 to 9. So you get 39 which is just one third. So now we don't be well, now that we know what B is and we know what a is Remember, let's see what's equal to negative six a. And we know what a s a negative six time to ninth and that comes out to be negative for third. So now we have a B A B N c. And to find what D is we can There's a There's a number of ways to do this. I'm just gonna do the simple one. So everyone which is a B plus a plus B plus C plus d, which is gonna be to ninth plus one third and then she is negative. So it's me minus four third and then plus D and we know that everyone is equal to zero. And if you saw for D, you got 7/9. So now we have salt for A B and A, B, C and D. And if you plug in those values into our original function, I mean to our original form, that is the cubic function.


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