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(3) Let G = Dv: Find H = [G,G]: bF What is IG / HI? Is G/ H cyclic?...

Question

(3) Let G = Dv: Find H = [G,G]: bF What is IG / HI? Is G/ H cyclic?

(3) Let G = Dv: Find H = [G,G]: bF What is IG / HI? Is G/ H cyclic?



Answers

For $g(u)=3 /(u-2),$ find and simplify $[g(x+h)-$ $g(x)] / h$.

I'm gonna do these next four problems altogether because they're very similar. And so what they do is they give us a function H that's a composition of two other functions and then it give us the value of the derivative and the function itself at various points. So we have a choice is F composed with G. We have F prime of two is six. F of one is four. G of one is 2 and G prime of one is -2. So they want us they want to know what H prime X equals one is so H prime of X is F prime of G times G prime of X. So we plug in x equals one. So that's f prime of G evaluated at one that X equals one Times G. Prime Evaluated at one. So G evaluated at one is too, so we get F prime evaluated at two. G prime evaluated at one is minus two. F prime evaluated it too Is six. So we get six times -2 is -12. So we haven't, we have enough information about the two functions and in fact we have extra information because we never actually used this, we have enough information to figure out. We don't know what the functions are, but we know enough information about the derivatives that enough points to figure out what the derivative of the composition of the two function is, functions are at X equals one. So the next problem is very similar, almost identical just with different values here. So H prime of X equals F prime, the F D x D F D G. And so G is a function of acts and then G d g d X. So H prime at one equals F prime Evaluated at the value of G when it actually cause one. So G f one is three. So we get f prime evaluated three. And then we need G prime evaluated at one and that's three here. So f prime evaluated at three is 4 Times three is 12. Now we just have a G equals G composed with F. And again, we're given enough information here about these two things about the these two functions and the derivatives at enough points. Because we want H prime at zero. So H prime of X equals G prime at a fair value at f of x times F prime. It acts so we plug in X equals zero. So we get G prime evaluated evaluated zero. F evaluated at zero is three. And then we get f prime evaluated at zero is -1. And so we get the minus sign out here and then G prime evaluated at three is zero. So in the end we just get that the the hdx evaluated x equals zero is zero. So we got that from knowing this this information again, we didn't need that that piece of information just these three. Now the same problem here, we have a church G composed with F. We have um extra information here. We have F and F prime at one and at two And then G&G prime at x equals -3. And we want to find a cheap primer to so H prime the hdx he caused the G D F and F is a function of X. And the Fdx. So access to so we plug into and F evaluated it too is minus three. So we have G prime evaluated at -3. F. Prime, evaluated it too is four. And so then we need, so we have four here, so we need G prime evaluated at minus three And that's here, that's three. So we get three. This is three times four is 12. So we actually didn't need any of this information here. Nor did we need um See we don't need G. The value of G At -3 that we didn't use either, but in the end H. Prime evaluated X equals two, is 12.

Being asked is what is the prime of one equal to? So the first thing that we need to do is take the derivative. So then per chain really have H prime of FFT and then times F prime of team. Okay, so now plugging this in, we have G prime of one is equal to H prime of F. Of one. Times have prime out of one. Okay. Well, you know that half of one Physical to three. So then this is equal to h. prime of three And then have private one is 6. So that's going to be times six year. And they gave us that H prime of three was equal to negative two. So it's negative two times six. So our final answer is -12.

So this question wants us to find the derivative of age at two. So here we are given the function of H. And we can use the power rule to get the derivative. So we have F prime G. X. Because that's an inside times insides derivative ogx. So now getting it at X equals two requires us to know G of two and G prime at two. And luckily were given both of those as initial conditions. So, And we're also given f of you. So we need to find the derivative of f of U. two as a function of you. And that's just to you. So this is able to F prime of G at two which is three. So f prime at three times G prime at two, which is a negative one. And now we can plug in three into this derivative we just found and that gives us six times negative one and that gives us negative six. So the derivative of H at X equals two is negative six.

Our goal is to find our prime of one. We have R of X. It's a composite function. So we're going to use the chain rule and let's start by finding our prime of X. So we start with the derivative of the outside function F with F prime of G of h of X. Then we moved to the middle function G, and we have its derivative g prime of h of X. Then we moved to the inside function H. And it's derivative is age prime of X. Now we want to substitute one in here so our prime of one would be f prime of G of age of one times g, prime of age of one times h, prime of one. Now we can start substituting some of the numbers in here. H of one is to we get that again and then we get h prime of one is for So when we substitute those in, we have f prime of G of to times G prime of to times four. Now we need G of to G of two is three and when you g prime of two g prime of two is five So now we have f prime of three times. Five times four. Okay, One more step to go. We need F prime of three f prime of three of six. So now we have six times, five times four and that's 1 20


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