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Bbwal [pd 6-7 Latfu) ens) plke 4p Icn #b . Uefy (tac = Edlee_Theorem aPplied Rolle'$ conclusion = [0,6] satisfying the Point € In , interval L b) Find t...

Question

Bbwal [pd 6-7 Latfu) ens) plke 4p Icn #b . Uefy (tac = Edlee_Theorem aPplied Rolle'$ conclusion = [0,6] satisfying the Point € In , interval L b) Find the _ to f(1}

Bbwal [pd 6-7 Latfu) ens) plke 4p Icn #b . Uefy (tac = Edlee_ Theorem aPplied Rolle'$ conclusion = [0,6] satisfying the Point € In , interval L b) Find the _ to f(1}



Answers

Let $f(x)=p x^{2}+q x+r$ be a quadratic function defined on a closed interval $[a, b] .$ Show that there is exactly one point $c$ in $(a, b)$ at which $f$ satisfies the conclusion of the Mean Value Theorem.

Questions because you fight. We need to verify rulers. Sure, Um, and this particular interval. By finding that F is equal to a FB, there is the starting point, the standing point. So let's find a way, which is a. For one or two. It will be one or two plus 1/1 or two, which is nothing but one or two plus two, which is five or two on. Let's find a B, which is F two. It is two plus one or two on two plus one or two. It's nothing about five or so. Both are equal, so we can apply the rollers, Children, Uh, for the roller student, we need to differentiate this first. So differentiating X is one differentiate differentiating one of our eggs will give minus one over extra square. So we have F dash C is equated to zero. It means that we have one minus one of us. See square zero. This means that we have one oversee square as one. This means that we have the C square as one. This means that CS plus minus one, but rule for rulers still on the value C should be between a and B, so she should be positive. So C is equal to one, which is in fact between one or 2111 of what you want to as well. So it means that the ruler student is satisfied on the solution, for that is equal to one for which F for which the ruler store, um, is getting satisfied for a given function on between a given interval of one or two people.

About questions. 67. We need to first verify that we can apply rules you run toward by finding every n f B, which is the which other starting at in points and proving that they are saying so. If they would be F three, which is three square over 18 threes, 20 for minus 15 which is nothing but nine or 19 which is nothing but one on Let's Find F B, which is a five as well, which is five square, where eight times five is 14 minus 15 which is 25 or 25 which is again one. Since both of these are same, we can apply the rollers Children, so it states that we first have to find a derivative so derivative way have to use the question. True. So we have denominator square. The normal Attar has an eight x minus 15 As Attar's differentiation off numerator is extra square will be two x minus numerator differentiation of eight X minus 15 which is denominator is just eight. We find we quit of dash C 20 now, so we have eight C minus 15. When we quit 20 Obviously the numerator will be zero because the denominator does not have any significance here. So we could skip writing the denominators. Let's open up the brackets here too. So we have 16 sees where? Minus 30 C minus hcs where? Zero. This means that we have eight C square minus 30 C is zero we can take foresee out. So we're left for two C minus. In fact, we cannot take four c out. We can take to see out if we take to see our real effort for C minus 15 is equal to zero. From here, we get to values off C 10 and one is 15/4 0 is not between three and five, so that is rejected. But 15 or four is certainly because that is three point something on three point. Something is definitely between three on, uh, five. So we'll take 15/4 as a solution. So finally, we can say that F Dash C is equal to zero at C equal to 15/4, which is between three and five. So this rules the roller still

All right. So you have FX is X to the two thirds on the interval, zero to one. We want to find the value of C satisfying the conclusion of the mean value to him. So we need to review them. It's going to be to over three Q brood of X, some derivative, right? And so, if we set of prime of C equal to half of one minus F Syria over one minus zero, we have to or three cube root of X equals Okay, one minus. Syrah won Monastir. That's one. So then, if we flip this, we have three. He root of X over two equals one whore. Cuba of eggs is equal to two thirds. If I multiply by two invited by three. And besides, so then exit by cube those sides I get eight over twenty seven. So I wantto just let c b eight over twenty seventh, and that would be the condition. Or that would be the point and open interval zero one that satisfies the conclusion of the mean value. There

It was verified that the function G here satisfies the hypotheses of the mean value theorem on the closed interval from 0 to 8. So to do this we must prove two things. One that f is continuous on the closed interval from A to B, and to that it is differentiable on the open interval from A to B. So first, let's consider if it's continuous. So when we look at the graph here and here in red, there are no visible gaps, holes or jumps of any kind. And the function is simply a smooth flowing connection of points with no breaks in it. So first her part, the CIS is indeed fulfilled. It is continuous. Now we must see if it's differentiable on the open interval for me to be and to be differentiable, the function must be first of all continuous and we've already proven that. But it must also have no sharp corners or kinks of any kind. It must be a smooth flowing line which it is here. So we've proven that it is differentiable on the open interval, not too close interval because the graph given does not continue the graph past the 0.8 or pass the X value is zero x value eight x value zero. So it would not be differentiable at the end here because it would be an abrupt stop. There would not be and you function to the right of the X value eight. But we don't have to worry about that because it's an open interval. So both things have been fulfilled. And now we must use the mean value theorem that estimate some values for C first in the interval, 0 to 8 and then in the interval of 2 to 6. So first we'll do 0 to 8. And so if you look at the right hand side of the mean value theorem equation here, this is simply the formula for the slope between the two end points, you can see a difference and why over difference in X. So if I draw a line between our two endpoints this line, the slope of this line will be this part here. Want box it twice. It will be this function here or this value here and so to find. See, we have to find a point an X value at which the function has a derivative or slope equal to the slope of this line. The slope of the tangent line, I should say so we can do some estimating here we can see. We could say about right here. You can see the tangent line. If I could draw better, would be approximately parallel here. And then we can also have one about here where the tender line is about parallel. And so if we estimate we could say about 6.1 or 6.2 for this point and then the X value here would be about, I'd say 1.9. Or or we could just say to, I'll say 1.9. So let me get these lines out of the way. So for B, we said 1.9 and 6.2. Now for C, we have to consider the interval 2 to 6. So let's get our points on the graph here, points on our function, and once again we'll draw this line, which will have the slope equal to the right hand side of this equation. Here, there we are, and then we have to find points within the interval. They have a tangent line parallel to this line because of the tangent line is parallel. That means the derivative will be equal to the slope. So once again we can say its one about here, and we'll say We have one about here. And so we'll say, for C, we dropped this one down as well. So for part C, we'll say, Call that three point four and we'll say five points, five point two 5.1. No, these are simply estimates. And because I've recreated the graph here from the book, your answers may be slightly different. But not to worry, because these are simply estimation, so as long as you're within a few 10th, it should be all good.


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