Question

attached

Answer 1

Find (f+g)(x) and (fg)(x)
Find f(g(x)) and g(f(x))
Find f-1(x) and g-1(x)

Answer 2

sup you!

Answer 3

hey! sorrry we can do this quick lol :D
f(x)=5/x-2

Answer 4

g(x)=x^2+1

Answer 5

no problem, I have things to do as well...so let's do the first question f+g what do you think it is?

Answer 6

((5/x-2)+(x^2+1))

Answer 7

yes exactly! but just for good measure put them under a single denominator

Answer 8

so (5+x^2/x^2+1)?

Answer 9

ehhh no, I meant to add them together but now that I think about it, it's fine to just put it as: \[(f+g)(x) =\frac{5}{x-2}+x^{2}+1\]

Answer 10

so forget that I said to put them together...

Answer 11

ok so whats f*g?

Answer 12

alrigjt

Answer 13

((5/x-2)(x^2+1)

Answer 14

yup or more easily seen as: \[(f \times g)(x) = \frac{5(x^{2}+1)}{x-2}\]

Answer 15

make sense?

Answer 16

yes got it

Answer 17

alright so whats f(g(x))?

Answer 18

i forgot how to put this

Answer 19

remember that when you write a function you write is as "f(x)" right, the "x" is there because the function is in terms of x. Now that we are replacing the x with g(x) we need to replace the "x"'s in the function with what g(x) is equal to. Does that make sense?

Answer 20

umm yesss so f(x^2+1)

Answer 21

yes exactly! That's what f(g(x)) is really asking for! So..what would f(x^2+1) be equal to?

Answer 22

5/((x^2+1)-2)

Answer 23

yes!!! Which simplifies down to?

Answer 24

5/x^2-1

Answer 25

wait

Answer 26

did you forget the coefficient with the x^2?

Answer 27

g(5/x-2)
5/((x^2+1)+1)and what do you mean

Answer 28

g(x) is not x^2+1 right? it's 2x^2+1 so it shouldn't be 5/((x^2+1)-2) but rather 5/((2x^2+1)-2) see the difference?

Answer 29

oh ok yes

Answer 30

alright good so if you simply that it would be?

Answer 31

so then f+g i s 5/x-2+2x^2+1

Answer 32

and yes! But you should use parentheses otherwise it's hard to tell...so (f+g)(x) = 5/(x-2)+2x^2+1

Answer 33

and fg which we previously had as 5(x^2+1)/x-2 would be 5(2x^2+1)/x-2
which won't it simplify to 10x^2+5/x-2?

Answer 34

and 5/x^2+2

Answer 35

yes completely correct!

Answer 36

wait what about 5/x^2+2?

Answer 37

g(5/2x^2-2)
5/((2x^2+1)+1)
5/(2x^2+2)

Answer 38

ummm I'm a little confused at what you are doing right now well actually starting from your post: "and 5/x^2+2"

Answer 39

it would be 5/2x^2+2

Answer 40

what would be?

Answer 41

g(f(x))

Answer 42

oh! ummm no lol.

Answer 43

ohok

Answer 44

so just like f(g(x)) where you put g(x) where there was any "x" in f(x) do the same with g(f(x))...does this make sense?

Answer 45

it's just opposite now...

Answer 46

2(5/x-2)+!

Answer 47

+1

Answer 48

yes! but remember that x is squared right so...?

Answer 49

by x I mean in g(x)

Answer 50

2(5/x-2)^2+1

Answer 51

right!

Answer 52

or most clearly written: \[g(f(x)) = 2(\frac{5}{x-2})^{2}+1\]

Answer 53

make sense?

Answer 54

wait so next distribute the ^2 to the 5 x and -2 in prans or distibute the 2 to the 5 and x-2?

Answer 55

yess

Answer 56

\[(25/2x-4)^{2}+1\]

Answer 57

well to expand the ^2 out we just simply multiply by itself so:
\[(\frac{5}{x-2})^{2} = (\frac{5}{x-2})\times(\frac{5}{x-2})\] make sense?

Answer 58

yes

Answer 59

oh you mean the 2 in front of everything?

Answer 60

yes

Answer 61

ok so we can't actually deal with that without first dealing with the ^2, because of the order to operations, do you remember this?

Answer 62

yes pemdas

Answer 63

yes so exponents are before multiplying right?

Answer 64

correct

Answer 65

so we deal with the squaring first. So squaring 5/(x-2) gets us what?

Answer 66

25/x^2+4

Answer 67

oooh close so (x-2)(x-2) doesn't equal x^2 +4 do foil or what I showed you last time.

Answer 68

x^2-4x-4

Answer 69

25/(x^2-4x-4)+1

Answer 70

yes beautiful so it now we have:
\[2(\frac{25}{x^2-4x-4})+1\]

Answer 71

you forgot the two in front remember?

Answer 72

oh yeah

Answer 73

ok now distribute the 2!

Answer 74

(100/2x^2-8x-8)+1

Answer 75

no so when you multiply a fraction things that in the numerator distribute to the numerator and things in the denominator distribute to the denominator. So 2 is really like 2/1 right so nothing really happens in the denominator...Does this make sense?

Answer 76

oh ok so just 25 changes tp 50

Answer 77

exactly!

Answer 78

so what's g(f(x))?

Answer 79

50/(x^2-4x-4)+1
50/(x^2-4x-3)

Answer 80

ooh but you can't just add numbers like that right? You must find the LCD to do that. Do you understand this?

Answer 81

oh

Answer 82

but it's really simple (honestly I would have just left it as 50/(x^2-4x-4)+1) but since you want to combine them I'll show you:
\[g(f(x))=\frac{50}{x^2-4x-4}+1\]\[g(f(x))=\frac{50}{x^2-4x-4} + \frac{x^2-4x-4}{x^2-4x-4}\]\[g(f(x))=\frac{50+x^2-4x-4}{x^2-4x-4}\]\[g(f(x))=\frac{x^2-4x+46}{x^2-4x-4}\] does this all make sense?

Answer 83

oh ok yeah sorry by internet went down

Answer 84

no problem, I was just laughing at one of the question someone asked on this site...

Answer 85

anyways does that make sense?

Answer 86

what question
and yes i got it
the one became the x^2-4x-4/x^2-4x-4

Answer 87

the question posed in the math section: how to have sex...lol.
And yes good!

Answer 88

ok now what about f^-1(x)?

Answer 89

omg lol thats too funny! crzy people
and would ti be f^-1(x)= -5/(x+2)

Answer 90

yup and no! yes that's funny but no the inverse is wrong...just switch the x and y in the original function and then solve for y. Show me btw.

Answer 91

y=5/x+2)
-2 -2
y=5/x

Answer 92

well this the reasons why you should use parentheses...you can't just subtract 2 since it is part of the denominator. furthermore the original function is y = 5/(x-2) right?

Answer 93

yes

Answer 94

would you like me to show you?

Answer 95

yes

Answer 96

ok so the original function is this:
\[y = \frac{5}{x-2}\]we switch x and y right? so...\[x=\frac{5}{y-2}\]now we solve for y! So our next step is to move the whole denominator to the other side:\[x(y-2) = 5\]does this make sense?

Answer 97

gpt it

Answer 98

ok so now we should move the x to the other side to isolate the y:
\[y-2 = \frac{5}{x}\]then we move the -2 right? so:\[y = \frac{5}{x}+2\] does this all make sense?