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C2Let g() = f()dt. Using ' the figure above; find(a) g(0) =(b) & (3)(c) The interval where gis concave up:<x <(d) The value of xwhere g takes its max...

Question

C2Let g() = f()dt. Using ' the figure above; find(a) g(0) =(b) & (3)(c) The interval where gis concave up:<x <(d) The value of xwhere g takes its maximum on the interval 0 < x < 8.

C2 Let g() = f()dt. Using ' the figure above; find (a) g(0) = (b) & (3) (c) The interval where gis concave up: <x < (d) The value of xwhere g takes its maximum on the interval 0 < x < 8.



Answers

Let $g(x)=\int_{0}^{x} f(t) d t .$ Using Figure $6.33,$ find (a) $g(0)$ (b) $g^{\prime}(1)$ (c) The interval where $g$ is concave up. (d) The value of $x$ where $g$ takes its maximum on the interval $0 \leq x \leq 8$.

All right. So today we're going to be looking at this function G of X equals the integral from zero to x of T t. T. And we are also given this graph here of F. So we're gonna be finding things like local maximum local minimums. Um, and to start off, we're going Teoh, remember that by the fundamental Fierman calculus, we know from this that g prime. So the derivative of G is going to end up Equalling yes of X, which luckily we have a Zagreb. So this, um, of here I'm just gonna right that this is g prying eyes. Your reminder, because getting it kind of confusing. So to find the local maximum of G, we're gonna be using the graph of the derivative G, which is F so the local maximum and men's air going to occur where the function F of X equals zero. So where the derivative equals zero? So where does that happen? Well, we can just look at the graph here, and wherever it crosses, the X axis is going to either be a local maximum or a minimum. So to figure out the difference, we're gonna look at the slopes of the graph. So it a maximum is going to be when you go from positive to negative. So up here we are positive. You're in a positive zone and we go down through two into a negative area. So that makes to a maximum of G. Similarly, from down here were negative. We go through four and end up positive, so four is going to be a local minimum of G. Also, we have six is a max cause you go from positive to negative and eight is gonna be a minimum because you go from negative to positive. Awesome. So we already entered the first part of the question. Next, we need to figure out the absolute maximum g. So the absolute maximum is going to be the highest point of the whole. Graf local that we're looking at one specific area that absolutely we're looking at the entire gruff. So to consider this, we're gonna go back to what we're given here of what g of X is, and we're gonna plug in the values, um, of two and six to figure out which one is the absolute maximum. Now, just a side note. Um, what about X equals 10 because 10 clean 10 is also where we cross the xx zero. So why is it not a maximum in? Well, the function is not actually defined past the point of X equals zero. So we don't actually know what happens on this other side here. Therefore, we can't conclude anything about the points. We're going to just leave it off for now. So just looking at two and six, we're gonna consider what G of two is so g of to is going to be the integral of 0 to 2 of f t D t. Which is gonna end up being the area under the curve here because we're going from 0 to 2. So for looking at G of six, we're looking at the integral area under the curve from this point. So from looking at the Inter grows, we can conclude that G of two is going to be bigger. Well, that was huge. I'm sorry. Do you have to Is gonna be bigger than G of six. So X equals to is going to be our absolute maximum of G next. We want to determine when G is calling Cabe down socom, Kate, come Cavity is determined by the second derivative. Yeah, and for it to be conch it down. We know that second derivative of G has to be less than zero. It needs to be negative. So when is second remunerative? Negative? Well, it's going to be lesson zero when the first derivative is negative slope. So this happens when the slope of G is prime is native, and we already know that G prime equals f on our graph here. So we're going to figure out at what points is on the slope is going to be negative. And that is gonna be our intervals for come cavity. So I'm just gonna follow this black line here for positive positive positive switches up here and to negative until you reach here goes back positive that were negative again. And then we're negative. So Khan cavity is going to occur on the interval of one, 23 five, 27 and from seven to Oh, excuse me with from sorry from the line. We're looking at nine to 10. Awesome. So those are there are intervals where we're conch it down. Lastly, we want to grab the original function So we started with the function of the graph of the derivative, and now we need to go backwards to function or the graph of the original function. So because of the work that we just did, we know the intervals where it is con cave down. So calm cave down is gonna look something like, Whoa, not like that, something like that. And we also know the absolute maximum. And we also know the maximums and minimums. So I'm going to go in here, and we know that there is a maximum at two. There is a minimum for there's another maximum at six and another minimum at eight. We know it is Kong Cave down from 1 to 3 from the 5 to 7 and from nine to 10. And then from here, we're just going to connect through those points that we already know. And there you have it. That is our graph of the original function G of Lex

And this question were given a graph of a function F. And we're told that G is the interval of this function. And we want to find out a couple of calculon questions, the first one being at what X do we have the local max admit this means that we need to have G prime to be zero. So that means we need to find obviously G prime G prime of x equals just by FTC one. It's just going to be the function. So we can just look at the values of the graph and we need to find out where, where the sign changes. Remember that if the sign changes from negative to positive, we know that this function will have a local minimum at that point. And if it goes from positive to negative we'll have a local max at that point And we can see from the graph that the local minimum occur at three and 7 simply because of the changing signs. And by the same logic, if the signs change in the other way It will be at one and 5. The second part of this question asks us where is the absolute max? So you need to chest that means we need to look at the size of the integral. Uh huh. And where it increases the most and we noticed that with these functions, the more we go, the more we go to the right the more area we get under the curve. If you think about it. So if you think about it, we move remove in some amount higher upwards. So we can say that the maximum value is at the end of the interval. That is at x equals nine. The third question is where is G concave down in order to do that. We need to look at the con cavity so we know that G double prime is equal to the first derivative of the function that's printed in the book for this. We need to look at where the function is, where the tangent lines are zero. Yeah, where there's slope equals zero. That's what we want to find. Oh yeah. Or alternatively we remember that the parts that are concave down from just from observation are those that are decreasing those in which the function is decreasing. Mhm. So, and this happens on the following intervals. Yeah, on that interval on that interval and on that interval. And the last thing we want to do is we want to sketch the graph. Mhm. Community the graph of G. And we can do this by keeping in mind a couple of things. The first thing we should find is one point on the graph, it's namely Gm zero and we'll do that right now. And since it's defined as an integral, we know that if I integrate over a point and I end at the same point It will be zero. Yeah. And given this, I know the values of, I know where the local and absolute maximum will be. So now we get to pretty much we get to graph this function. We know that it will be increasing forever. Not forever. It will be, it will be something like this. So pretty much like the sign curve that we're given here. They're not really a sign curve, but it will look something like it looks something like this. It this is solely based off of the information given and the question that we just solve for that. So I do this question.

In this question were given a graph of F and we're told that G is the integral of F. And we want to find a couple of calculon questions out the first one being where the local min and max remember that local men and max is involved G prime. So we can say that G prime of X. Remember that that's going to be equal to just the function by the fundamental theorem of calculus. And let's remember the set, we have to set G prime equal to zero. And remember that if if F goes from negative to positive, remember we have a local men and if it goes from positive negative, we'll have a local max at those points. So clearly clearly as you can see, we'll have local minima at X equals four and x equals eight and we'll have local maximum at X equals two and six. Yeah. This is purely by observation. The next question I want to find as you want to find the absolute max of this function, you know this? Well, how can we do this? Well, if we look at the function, we know that The function G will be increasing the most until around X equals two. Mhm. And all of the humps that all the intervals of increase or decrease. They happen They secured less. So we'll assume that the absolute max is going to be x equals two because it's a little bit hard to approximate it. You could you could be a little bit, you can be a little bit let's show a little bit farther. But we'll call, we'll call X equals to our absolute maximum. Just by approximation. The third part is we want to find where it's concave down where the function be, concave down, remember that con cavity involves G double prime and we can just look at the first derivative effects of the function F. So you have to look at where it's zero. I used to look at where at zero. And remember that the concave down parts are where G prime is decreasing. So. And by estimation, yeah, it's the same space. By estimation, We want to we're going to attempt to say that this happens when X is in between .8 and 2.8 looks like. Uh and also when X is between 4.8 and 6.8 and also from 8.8 to 9.8. The last thing we want to do is to sketch the graph. The first thing we should do is we should find a point on the graph, namely GG 00. And since it's defined as an integral, it'll just be zero because of how the integral will work. So our function from the information that we found earlier. Well, look something like this. Mhm. That's how you do this question

No basis for this problem. They give you an interval Integral function F of X. Like also the integral of team Ministry over T Square plus seven DT from zero X and for this problem has three parts A, B and C and for part A. They want you to find the value of X where, um f of X has a minimum value whole, the whole middle value. So if we were to find the minimum value of this function, what we need to do is to find the critical points right and to find a critical points, we need us take the derivative of the function and said the equal to zero because at the minimum, we assume that we can assume that the middle of this Qianjin line or a slope would be able to zero. So since we're dealing with an integral function, but we don't take the derivative of our function, we've yet X minus three over X squared plus seven. All right, so if we were to set this equal to zero, if we were to set this equal to zero, the only values where this would work would be X equals a three since three minister would become zero and you get snow us? Ah, for whatever value, actually, choose that the nominator will always be a positive value. There is no way you can get undefined for denominator. So our only value possible value is X equals 23 Alright, so that's four party. Uh, now, for part B, they want you to find increasing and decreasing intervals for the function every backs. All right, so what we need to do here is create a table. So I'm gonna rewrite the derivative over here. Because if F prime of X is positive, then that means the function is increasing at that point. And if it is negative than the function is decreasing at that point on the left arm of X equals zero, then the function is either either increasing or decreasing missile zero. All right, so this is the derivative of the function that's will be a seven and hard to create my table. We know that at X equal to three, the function of X equals zero. So we're gonna take note of that and create a table or when X is less than three on. All right. I'm of X here. So we're gonna have a table like this or access less than three in another part for when X is greater than three. All right, so when X is less than three, if we were to plug in something like negative for would obviously get negative value and the numerator and a positive value in the denominator, So it will be a negative value and for a value x greater than three. So let's save replied in positive form we get for my three, which is one and normally we have a positive number because execrable seven is going to always be positive. So we have a positive for interval exes greater than three. So using that information, we can write for increasing, we're increasing. Uh, the interval for when the function is increasing is when, uh, X is greater than three. So from three to infinity, because that frame of X is positive. So we're decreasing for decreasing. It will be from negative infinity 23 because our current backs is negative on that interval. All right, so box these answers. All right, So now moving on to part C currency is dealing with the Cong cave con cavity of frame. So what? So I have stubble prime. We know that as double prime of acts deals with the cavity of a function and the F double prime is positive. That means the function is called gave up. And if F double prime is negative, then it's con cave down. All right, so and if at the crime, because zero, then that's an inflection point, meaning that it's changing from comedy. Our son we need to do is to take the second derivative of our function. If we do that, we're gonna need to use ah product not sorry the quotient room. So using the question will get X squared plus seven minus two x times X minus three all over X squared plus seven squared. And if you were to simplify this, we get negative of X squared minus six X minus seven. All over X squared plus seven square. Now we can factor the numerator could finally get that's your negative. X plus one x minus seven divided by X squared plus seven squared. All right, so now what we need to do except is equal to zero to find. When I told prime of X crosses the X axis. And we know that from two expressions on the new Raider are zeros are at X equals to one, and X equals two X equals negative one. Sorry and X equals two Brussels seven. Positive seven. All right, so now for to find the conduct find contracted E we again have to create another table, Uh, this time with f double prime of X. And first we're gonna have the values of X that are less than negative one. And next, we're gonna have the values of X that are greater than negative one. But last in seven, you're drawing Z's 77 and finally will have X values that are greater than seven for our last interval. So you have double prime of X. If we were to plug in a value of X and a smaller than negative one, we get ah, negative here a negative here and they have here. And if you multiply them all together, we get a negative. So I'll be negative if we plugged in. Our value lies in between 81 and seven. Let's say zero we get native or negative sign here. Uh, positive one here and a negative seven here. So at the end, the natives cancel out to be a positive value. And finally, if we choose our value of X, that is greater than seven. What happened? We still have our negative symbol that's outside Andi experts. One would be positive, and x 27 will also be positive. So this will again be made of So we can draw our conclusions and say that the function is going to be conquered up on the interval from negative 1 to 7. Because after all, crime is positive there in that interval, and the functions gonna be conclave down when, um on the interval from negative infinity to negative one union and from seven to positive infinity because after all, prime is pause is native on that interval, that is, it will do your answers for all three parts of the problem.


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