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Suppose g(u) ViFedl Fiud the value o (g-'Y (0) ifit exists. If it does not exist, explain why: Hinb; Fundamental Theorem of Calculus...

Question

Suppose g(u) ViFedl Fiud the value o (g-'Y (0) ifit exists. If it does not exist, explain why: Hinb; Fundamental Theorem of Calculus

Suppose g(u) ViFedl Fiud the value o (g-'Y (0) ifit exists. If it does not exist, explain why: Hinb; Fundamental Theorem of Calculus



Answers

Suppose the function $f$ is defined on $[0,1],$ as shown in the figure.
$f(x)=\left\{\begin{array}{ll}{0,} & {x=0} \\ {\frac{1}{x},} & {0<x \leq 1}\end{array}\right.$
Show that $\int_{0}^{1} f(x) d x$ does not exist. Why doesn't this contradict Theorem 4.4$?$

In this problem, we are asked Explain why we cannot apply rolls serum to a particular function in this particular function were given F of X is one minus the absolute value of X minus one. And we're on the interval 0 to 2. So for roles serum, we, uh, have to meet certain criteria in order to apply it. So the first criterion criterion is that f is continuous on the closed interval in this case will be easier to to We also need f to be differentiable on the open interval, 0 to 2 and then we need f have a two equal f b. So we need f of zero to equal f of two. And in both of these cases, we get that this is equal to zero. So since both of these equal zero, our first criterion of f of A is equal to F B, we can check that off. We then look to see if f is continuous. So the absolute value of X is continuous. We shifted over to the right by one. It's still continuous. We flip it over and then shift it up still continuous. So we have that it is continuous on the interval, 0 to 2. So we then need to check. Is it differentiable on 0 to 2? And if we're not familiar with the shape of the absolute value, this is this is going to be the key. So f of X is one minus the absolute value of X minus one. You can think of the absolute value of X minus one as what is the distance? X is from one. So when we're at one, the distance is zero. So we have one minus zero, which is just one right. We don't say when I'm at zero. How far away is zero from one? It's one unit away from 11 minus one is zero. So I have a point at the origin at two. How far away is to from one? It is one unit away from 11 minus one is zero and we're not sure about the shape yet. We can fill in more points, right? So X is one half. How far is one half from one? It's one half unit away. One minus one half is one half similarly three halves. So we get the shape kind of like this and we look at this peak. This is what we call a sharp turn. And if we have a sharp turn in the graph, that means it's not differentiable at that point. So f is not differentiable at X equals one. One is in our opening of room interval from 0 to 2. Therefore, it's not differentiable. Yeah, and we cannot apply Rolle's theorem.

In this problem, we're given a function that where is X when it's is between zero and what, what and when. X is one physical zero, and we are trying to the terrible weather wolves. They're a slice for dysfunction. We don't have a functional, effective in. Points are equal, and the bat signal zero over rolls. Tara requires the function to be continuous. So now let's Evolet deep function, actually at the cut off point. So when exes they would want the full D function, it top when X is one f respond to one and from the bottom one when X is one f, it's fun to be zero, since those are not equal at the cut off point. The function F is not working years, and since it is not continuous, both era does know what

We're doing used function X is equal. Teoh that takes he cools 60 Where here It's less than you T X here to return to Germany. What African instrument? There's a derivative into Europe according to rules. Spc. Better of equal Teoh prime ex people's here according to her rules here. All right, with statements false since Fx's neck.

Can we apply rules theorem to the function? F of X is equal to the absolute value of X for any value X that is greater than zero. Okay, so let's look at this we have here. We'll just call this negative A We'll come over here and call this a Okay. So if we have any value of A that is greater than zero so not right here, can we apply rules theorem to that? And the answer that to that is no. Because you remember Rolle's theorem says that between the two book points that we're looking at, the function has to be continuous. Is our function continuous? Yes, it's continuous. We go from here. Here we go. Continuous, We just keep going. Okay, so we are continuous. The second stipulation says it has to be differential between those two points. Go here. We hit a kink in the graph. You cannot differentiate right here because there's a kink in a graph. It is not a smooth, continuous curve. Therefore, f of X is equal to the absolute value of axe. You cannot apply rules theorem to


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