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TH [G20Final Kenlon (Pt} Ler 4 En 18417.6,2 Tour Werl! 77235,71 Alxo, Ist tr funrIlont /.47800-4 "lncub} {(9,31(,7,(,21,6.5, (LSi6 W 4 2.(4.5.",(7,91 rude...

Question

TH [G20Final Kenlon (Pt} Ler 4 En 18417.6,2 Tour Werl! 77235,71 Alxo, Ist tr funrIlont /.47800-4 "lncub} {(9,31(,7,(,21,6.5, (LSi6 W 4 2.(4.5.",(7,91 rudefle))[email protected]))_(s,ftos) (3ftollz {(242),63,2),s,5),C (7,38 6) Finx"f = {4-9521,(u94))_(a%e) ln.94) (Jts) (2,949)3 {t0,t) , (u, 8),(,2),(a).(t,t),(2,2) } hich J(4ny) of the functions Inirctit? Juatit-vnurinatert NO , Sinte flv) = 5 f(e) Yes Which ( any) of the functions 3 JWjec[4un Jutilvquur 4ntnerl 9es, since f(A) No , Sinke 9

TH [G20 Final Kenlon (Pt} Ler 4 En 18417.6,2 Tour Werl! 77235,71 Alxo, Ist tr funrIlont /.47800-4 "lncub} {(9,31(,7,(,21,6.5, (LSi6 W 4 2.(4.5.",(7,91 rudefle))[email protected]))_(s,ftos) (3ftollz {(242),63,2),s,5),C (7,38 6) Finx"f = {4-9521,(u94))_(a%e) ln.94) (Jts) (2,949)3 {t0,t) , (u, 8),(,2),(a).(t,t),(2,2) } hich J(4ny) of the functions Inirctit? Juatit-vnurinatert NO , Sinte flv) = 5 f(e) Yes Which ( any) of the functions 3 JWjec[4un Jutilvquur 4ntnerl 9es, since f(A) No , Sinke 916) TA How nam Larni Juac(arie[c (ncrc Irolt lel 4092_Justity your answer! How ntany dillen- swrjectie unclols rom Aoar Justify your Eoswer! E) How mzny dilkxnf injective lunchansarrom Lh4? Justity yoer zn972r! 161 = 427 I6 and funcnen Muft 0 € Si ngi e va IUC h) How many difkou Elatians 4c thcre fron Justify yowr onswer? 32 pord



Answers

Compositions of Functions In Exerci $\boldsymbol{f}^{\circ} \boldsymbol{g},(\mathbf{b}) g^{\circ} \boldsymbol{f},$ and, if possible, $(\mathbf{c})\left(\boldsymbol{f}^{\circ}\right.$ $$f(x)=\sqrt{x+4}, \quad g(x)=x^{2}$$

So they want us to use these two charts topless. Find these compositions for G composed with f. So remember, all this notation right here means is that we're gonna write g of f of X. So each of those X values, we first plug it up. And then whatever that output is, then we plug into G. Let's see. So we start with zero. So that outputs zero. So then we come over to zero and then that outputs pie. How so? This 1st 1 is going to be pie, huh? Now for pie six. So we person played that in tow F so that gives us 1/2 become fined 1/2 over here, and then that outputs pie third. Then we do pie fourth, so outputs route to over to. And then that should give us this one right here, which would output pie fourth. And then we do pie third. So that outputs through 3/2. It's over here in G. That's gonna output pie six. And then lastly, we do pie half soap. I half outputs one and then over here, one, I'll put zero for G of why

Were given to functions that were composed them. And the first composition we'll do is we'll put F of G, so G will go into F. This will go in over here, so we have X over two plus three, and that's the answer for that one. Gff means this is going to go in Maddox, this one right over here, and we have the square root of X plus three Over two, and that's done. And up of five means Radical eight, and if you're asked to simplify it reduces to that, we get too radical too.

Going to compose uh these two given functions and the first one is F. Of G. So the G value is going in for F. So this is going in for that. So we have radical X squared plus one, which gives us X plus one. And now we're gonna flip flop that, we're going to take the F and put it in for G. So we're putting X square plus one over here. So I get this. And the last thing we're gonna do is F of five, we're putting five, and for F five squared plus one equals 26. And that's it.

We have the function capital F of X, which is defined to be f of f of X, where lower case affects is the function from exercise 47 and we want to find the derivative of capital f of X at X equals two. Let's start by just finding the derivative function itself. Well, we here we're going to be using the chain rule because we have a composition of functions. In fact, both functions are lower case F. But so what do we do by the chain rule? We look at the outermost layer, which is the function f and we take it's derivative, keeping the input value the same. And by the chain rule we multiplied by the derivative of the inner function which is of X so times f prime of X. Now all that's left to do is plug in to so we get f prime of and now from exercise. 47 were given that lower case F of two is one times f prime of two, which again from exercise 47 we are given is equal to five and yet again from exercise. 47 were given that f prime of one is four. So we have four times five equals 20. And now we're asked to do the same thing for this function little G of X, which is defined similarly as g o g of X. And this time we want to find G prime of three. So we do the same thing. G Prime of X is equal to a lower case. G prime of G of X times G, prime of X and we plug in three. So you g prime of and lower case G of three is too g Prime of three is nine n g. Prime of two is seven. So you get seven times nine equal 63 we're done.


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