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Since we ve required that f is three times differentiable, we know that f and it'$ first two derivatives are continous on1 [-1,1] From sub problem 3, we know t...

Question

Since we ve required that f is three times differentiable, we know that f and it'$ first two derivatives are continous on1 [-1,1] From sub problem 3, we know there constant, which we I call _ K,s0 thatIf() < K, If' (s) < K, If" (c)l <K Iel-1.1]From item we know thatf' (r)sin(nrr)drnTUse the previous sub-problemn to show that2K Iaal "Use the squeeze theorem to conclude that lim an =0 n_OC

Since we ve required that f is three times differentiable, we know that f and it'$ first two derivatives are continous on1 [-1,1] From sub problem 3, we know there constant, which we I call _ K,s0 that If() < K, If' (s) < K, If" (c)l <K Iel-1.1] From item we know that f' (r)sin(nrr)dr nT Use the previous sub-problemn to show that 2K Iaal " Use the squeeze theorem to conclude that lim an =0 n_OC



Answers

Show that if a function $f$ is differentiable at $x=a,$ then $f$ must be continuous at $x=a$ Hint: Write $$f(x)-f(a)=\left[\frac{f(x)-f(a)}{x-a}\right](x-a)$$ Use the product rule for limits and the definition of the derivative to show that $$\lim _{x \rightarrow a}[f(x)-f(a)]=0$$.

First problem we have Activex minus F of a equals two. Um, after the acts minus after a over x minus A I am the choir say based on this, we see that the function is differentiable at x equals A. Then it must be continuous at X equals A. Because if we can write it in such a way as this, we see that as X approaches a, this is going to go to zero. But if it's continuous, we see that the derivative exists so we can use the product rule and see that the limit as X approaches a um, is going to be zero in this case. So that means that the limit exists, which means the functions continuous uh

For the following problem, we want to let F and G be differentiable functions with H of X being equal to F of G of X, and then we're gonna let you equals G of A and V equal G Fx. So we want to show the limit as X approaches you. Uh HIV, Is he going to 0? So basically we want to show that um F V mhm for F of gva minus F of G of X is equal to E V. But we know that HIV is going to be um F a v minus F of you or review view. So it's going to be this right here divided by a b minus you. So we end up seeing that we're going to have this out front, B minus you. So sure enough, we would multiply those and those would be the same statement.

This question. We want to prove that if we take a constant and multiply it to f prime of a, then we would get something that looks like this. The limited. Each approaches zero of F of a plus, see the constant times H minus f a all over each. So let's look at the definition of F prime of A that's equal to the limit as age approaches zero of F of A plus, each minus F a, all divided by each. But what if we want to change what age is equal to? Well, we could just multiply a constant see Time's both of those H values. Now let's get rid of this. See on the bottom by multiplying both be left side and the right side of the equation by sea. And what that would get us is something that looks like this. See, Time's F prime of A is equal to the limit as H approaches zero of F of a plus C times H minus f of a all divided by h. So that's how we got our original equation that we're trying to prove is if we multiply both of the each values in the definition of a limit by a constant

Next to the four thirds hath one driven event zero but not too. And then next to the seven thirds has to derivatives but not three. And then find a general rule. Okay, so we're gonna start with Webex equals X to the 4th 3rd. So it's derivative four thirds X. To the one third. So F prime of zero equals four third time zero to the one third which is zero which is zero. And then the second derivative four nights X to the negative two thirds. So have double prime at zero equals 4/9 0 to 2 thirds which is zero which is for over zero which is undefined. So the second derivative doesn't exist but the first does all right. Next we have X to the seven thirds. So it's first derivative. Seven third X to the 4th 3rd. Second derivative, Yeah, 28 9 X to the one third. So after what prime of zero is zero. And then the third derivative. Yeah. 28/27 X to the minus one minus two thirds. So F double promote zero because the X. Is on the bottom. That's undefined or doesn't exist. Okay so now what we're looking for is a pattern so that what should KB so that it has in minus one derivatives and but it doesn't have in derivatives. So let's see what we have so far. We have X to the four thirds, say in minus one here. Extra four thirds had one derivative but not too next to the seven thirds has to derivatives but not three. So then X to the 13th would have three derivatives not four. So what's the pattern? Well the 1st 14 is three times one plus one and seven is three times two plus one and 10 is three times three plus one. Yeah so we want our exponents to be um three times in minus one plus one over three. That's K. Three m plus one. I'm sorry three three in minus one plus 1/3. So let's say we wanted something to have five derivatives but not six. Then in minus one would be five. So okay would be 16/3. So if X is X to the 16 3rd. So first derivative 16 3rd X to the 13 3rd. Next one 16 times 13/3 squared X. To the 13th. Next one 16 times 13 times 10/3 cubed X. To the seven thirds next one 16 times 13 times 10 times seven, three to the fourth X. To the four thirds. Can remember I wanted it to have five derivatives but not six. So this one will be 16 times 13 times 10 times seven times 4/3 to the fifth, next to the one third. And then you can see if you take the next derivative, the sixth derivative, you will have a negative power for X. And it will not be defined at X equals zero. Okay, so our formula works three in minus one plus 1/3. Do you want to simplify it? You can although it makes sense this way, because what you're interested in is in minus one. Really not in with three and minus three plus one over three, three in minus 2/3 in minus two thirds.


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