5

If f(x) =x - 9 and gx) =X 3x, complete parts (a) through (i): (a) (f+ g)x) = (Simplify your answer: Do not factor ) (b) (f - g)(x) = (Simplify your answer: Do not f...

Question

If f(x) =x - 9 and gx) =X 3x, complete parts (a) through (i): (a) (f+ g)x) = (Simplify your answer: Do not factor ) (b) (f - g)(x) = (Simplify your answer: Do not factor: ) 3 (c) (f- g) (Simplify your answer:)(d) (fg)(x) =(Simplify your answer: Do not factor:)(e)(Simplify your answer: Do not factor:)(f)(Simplify your answer:)(g) (f 0 g)x) (Simplify your answer: Do not factor:) (h) (g o f)x) (Simplify your answer: Do not factor:) (i) (g o f)( ~ 3) = (Simplify your answer: Do not factor: )

If f(x) =x - 9 and gx) =X 3x, complete parts (a) through (i): (a) (f+ g)x) = (Simplify your answer: Do not factor ) (b) (f - g)(x) = (Simplify your answer: Do not factor: ) 3 (c) (f- g) (Simplify your answer:) (d) (fg)(x) = (Simplify your answer: Do not factor:) (e) (Simplify your answer: Do not factor:) (f) (Simplify your answer:) (g) (f 0 g)x) (Simplify your answer: Do not factor:) (h) (g o f)x) (Simplify your answer: Do not factor:) (i) (g o f)( ~ 3) = (Simplify your answer: Do not factor: )



Answers

for the given functions $f$ and $g,$ find: $$ \begin{array}{llll}{\text { (a) } f \circ g} & {\text { (b) } g \circ f} & {\text { (c) } f \circ f} & {\text { (d) } g \circ g}\end{array} $$ State the domain of each composite function. $$ f(x)=2 x+3 ; g(x)=3 x $$

Okay, so our functions here are F of X equals two X plus three, and G of X is equal to three X. Uh So first one to find here F composed G of X. F composed G of X means F of G of X. Where we take G F Xr inside function and input it into F. So F of G of X is equal to well F of three X. That's equal to two times three X. And then plus three, which equals equal to six X plus three. So the F composed G of X is equal to six X plus three. And whereas the state uh the domain here, we can see this is just a linear function. Um And the domain is going to be the set of all real numbers. So we can say our or negative infinity to infinity. Right? The set of all growing numbers. Um Okay. And then for B um we have G composed F of X. So that's going to be G of F of X. So are inside function is now F. And put it into G. So we have G of two X plus three. That's equal to three times two X plus three. Which is equal to well six X plus nine. And again here this is a linear function, so therefore the domain is the set of all real numbers. And then four part C. We have F composed F of X. That's going to be F of F of X. So we take Quebec's input it into itself. That's going to be equal to F of two, X plus three. So that's two times X. So two times two X plus three we take F and we put it in two X into input. Then we have a three. We have two X. But X is our input of two X plus three. And then we have a plus three. So this becomes a four X plus six plus three which is equal to four X plus nine. And again here we see that the domain is the set of all real numbers. Excuse me. Okay. And then for party we have um G composed G of X O G composed G means G of G of X. So that is going to be equal to again, G of X is three X is going to be equal to G of three X which is equal to well we have three X three times X is our input of three X three times uh three X which is equal to nine X. And again we see nine X just over your function. So nine X. Domain is again the set of all real numbers. Okay?

Discretion of having at minus two or three. So that's 400 minutes. Be into a project that's being my ministry minus Yes, they put the energy that's being square minus Being thes permits me So when I was a minus, blessed be So that is gonna be spur Has to be this week.

So for this question, have the baggy or so let's go ahead and do so. So that's helpers. But some division there would have powerful be. So that's B minus three or Trilby. That's piece for minors. We cannot cancel anything personal with us, either. So let's just this is just leave it like that.

All right, composing two functions in various ways. First up is F F G. That's going to be F of two X squared plus three that's going to be equal to two x squared plus three squared and then add one. And you want to be careful here because you want to foil this in here so it's going to be too two X squared, plus three times itself, then add one. All right, So when I foil, I get four x to the fourth plus six x squared, plus another six x squared plus nine plus one. So combining my two like terms, uh, two pairs of like terms like four x to the fourth but 12 X squared, plus 10 of my final answer Looking at GF. I'm just reversing that process. That's going to be g of X squared plus one, which is two times x squared plus one quantity squared plus three. Again, Be careful the foil here, and I'm gonna leave the two out of it for just a second. Me two times a quantity x of the fourth plus two x squared plus one and then plus three. And then I'm gonna distribute this too in Teoh each term to get to X to the fourth plus four x squared plus two plus three. My final answer two x to the fourth plus four x squared plus five. All right, kind of squeezing these nff is going to be f of X squared plus one or X squared plus one squared and then at another one And in this case again, we got a foil here it's gonna be X to the fourth plus two x squared plus one plus one So we can just change it at the end to make that he too and finally g of G is going to be too RG of to x squared plus three. And in that case, that's going to be two times two x squared plus three squared plus three. We're going to foil this again. And luckily we've already done that up here. We're actually more over here when the plus nine and that's going to be two times the quantity four x to the fourth plus 12 x squared plus nine and then at a three. After distributing the to its eight x to the fourth plus 24 X squared plus 18 plus three, which will just make a 21


Similar Solved Questions

5 answers
Can the sp2 hybridization on molecule #2 be proven through electron configuration, and if 50 what is the configuration?=N;tvsen : Sp`041 2 P2pS : L P; ?5p?1J1 Sp?435;0 ~N_ Nitvzm: Sp}, Lt Kally Sp?AI Sp"41^ 3p111 2 p111 1 Sp Spy ? S: | 2 P: 435
Can the sp2 hybridization on molecule #2 be proven through electron configuration, and if 50 what is the configuration? = N;tvsen : Sp` 041 2 P 2p S : L P; ? 5p? 1J1 Sp? 4 35 ;0 ~N_ Nitvzm: Sp}, Lt Kally Sp? AI Sp" 41^ 3p 111 2 p 111 1 Sp Spy ? S: | 2 P: 4 35...
5 answers
Linear algebra 110 (Ji; 3)Consider the basis 5 = {. 12 / for R - where 7 = (-2.1) and V2 01.31 . and let T : R - R' be the linear transformation such that T (-1.2.0) and T(Vz ) = (0.-3.5) Then Ti-4-5 =(2-7.5) 0(-1, 8. 5 10)(1. -8. 10)None(-2.7.-5)
Linear algebra 1 10 (Ji; 3) Consider the basis 5 = {. 12 / for R - where 7 = (-2.1) and V2 01.31 . and let T : R - R' be the linear transformation such that T (-1.2.0) and T(Vz ) = (0.-3.5) Then Ti-4-5 = (2-7.5) 0 (-1, 8. 5 10) (1. -8. 10) None (-2.7.-5)...
4 answers
A paired difference experiment yielded the results shown below: nd = 50 Xd = 16.8 Sd = 8 Test Ho: Pa = 15 against Ha: Hd # 15 where Ha = (04q H2) Use a = 0.01. b. Report the p-value for the test you conducted in part a. Interpret the p-value_
A paired difference experiment yielded the results shown below: nd = 50 Xd = 16.8 Sd = 8 Test Ho: Pa = 15 against Ha: Hd # 15 where Ha = (04q H2) Use a = 0.01. b. Report the p-value for the test you conducted in part a. Interpret the p-value_...
3 answers
Question 11 ptsFind the Laplace Transform of e-t + etcosts + 1T)(s? + 2s + 2) S = 1 (s + D)(s? + 2s +1) s = 1 (s + 1)(s? + 2s + 2) s + 1 (s +s? + 2s +D)0
Question 1 1 pts Find the Laplace Transform of e-t + etcost s + 1 T)(s? + 2s + 2) S = 1 (s + D)(s? + 2s +1) s = 1 (s + 1)(s? + 2s + 2) s + 1 (s +s? + 2s +D) 0...
5 answers
Let V be the set of all polynomials of the form J=ar' +b Show Vis a vector space by answering the following If y1 and Yz are polynomials in the set; what property must they have ? Show the scalar multiplication is closed by showing that ky1 where k is a constant, has the defining property: Show that addition is closed by showing V1 Yzhas the defining property_Find counter example to prove that V is not closed to polynomial multiplication:
Let V be the set of all polynomials of the form J=ar' +b Show Vis a vector space by answering the following If y1 and Yz are polynomials in the set; what property must they have ? Show the scalar multiplication is closed by showing that ky1 where k is a constant, has the defining property: Show...
5 answers
Question 3(a) Suppose that Jin {0) =1. Find lim f(z). 2 fl) -5 (b) Suppose that ling 4. Find lim f (z). 1-2 1 2 42
Question 3 (a) Suppose that Jin {0) =1. Find lim f(z). 2 fl) -5 (b) Suppose that ling 4. Find lim f (z). 1-2 1 2 42...
1 answers
Let $X_{1}, ldots, X_{n}$ be independent exponential random variables each having rate $1 .$ Set $$ egin{aligned} &W_{1}=X_{1} / n \ &W_{i}=W_{i-1}+frac{X_{i}}{n-i+1}, quad i=2, ldots, n end{aligned} $$ Explain why $W_{1}, ldots, W_{n}$ has the same joint distribution as the order statistics of a sample of $n$ exponentials each having rate 1 .
Let $X_{1}, ldots, X_{n}$ be independent exponential random variables each having rate $1 .$ Set $$ egin{aligned} &W_{1}=X_{1} / n \ &W_{i}=W_{i-1}+frac{X_{i}}{n-i+1}, quad i=2, ldots, n end{aligned} $$ Explain why $W_{1}, ldots, W_{n}$ has the same joint distribution as the order statistic...
4 answers
A diagnostic message can be sent out over a computer network to perform tests over all links and in all devices. What sort of paths should be used to test all links? To test all devices?
A diagnostic message can be sent out over a computer network to perform tests over all links and in all devices. What sort of paths should be used to test all links? To test all devices?...
5 answers
Draw the Lewls structure and the Line-Anglestructure for each of the condensed structures shown below (& pts): (CH ) CHCH;CI (1 pt ) CH CH,CH,CH;CI (1 pt )(CH),CCI (1 pt )CH;CHCICH CHi (1 pt)
Draw the Lewls structure and the Line-Anglestructure for each of the condensed structures shown below (& pts): (CH ) CHCH;CI (1 pt ) CH CH,CH,CH;CI (1 pt ) (CH),CCI (1 pt ) CH;CHCICH CHi (1 pt)...
5 answers
F(x)f' (x)3++ +f" (c)-3-1 0 1 2 3 4
f(x) f' (x) 3 + + + f" (c) -3 -1 0 1 2 3 4...
5 answers
Qucstion 2Which molecule has both alkene and ketone functional groups?0cRemalning Time: 38 minutes; 52 secondsQucstion completion _luSType here scerchhiyt
Qucstion 2 Which molecule has both alkene and ketone functional groups? 0c Remalning Time: 38 minutes; 52 seconds Qucstion completion _luS Type here scerch hiyt...
3 answers
Ecous_Folomath csicuny eduwebwork2/Math232_22803_Foldes_F19/08.4 Taylor_polynomials/1 /key== bjBnbainu9glvbnGWG8YPcEoJUVdaKaRuser-kwilliams point)Calculate the Taylor polynomials Tz(c) and Ty(z) centered al Ifor f(z) = cos(z)Tz(r) must be of the formA + Blc 6)+c6 - &)?where AequalsB equals: CqualsandTs(c) must be 0f the formD + E(c 6) + F( 8)" + G(r 69'where Dequals Eequals: F equals: G equalsandNote: You can earn partial credit on nis problemhere I0 search
Ecous_Folo math csicuny eduwebwork2/Math232_22803_Foldes_F19/08.4 Taylor_polynomials/1 /key== bjBnbainu9glvbnGWG8YPcEoJUVdaKaRuser-kwilliams point) Calculate the Taylor polynomials Tz(c) and Ty(z) centered al I for f(z) = cos(z) Tz(r) must be of the form A + Blc 6)+c6 - &)? where Aequals B equal...

-- 0.018883--