Question

(4x+1)+(x^2-5)
= x^2+4x-4, right?

Answer 1

Yes :)

Answer 2

thanks a lot!

Answer 3

You're welcome! Have any others you want to check?

Answer 4

yeah

Answer 5

If you want to type them, I can check them.

Answer 6

ok i will thanks :P

Answer 7

(x^2-5)/(4x+1) for (g/f)(x) =
(x^2-5)/(4x+1), x=/= -1/4

Answer 8

So, g(x) is x^2 - 5 and f(x) is 4x + 1?

Answer 9

yeah

Answer 10

And they want you to find (g/f)(x) ? Then yes, that would be your answer.
Then, for the second part, are they asking you to find (g/f)(-1/4) ?

Answer 11

no, just the first part. thanks

Answer 12

Ok :) Got other questions?

Answer 13

yeah a couple

Answer 14

Ok, I'll wait

Answer 15

Thanks. For f(x)=4x+1 and g(x)= x^2-5, what is (f x g) (x). Would the answer be 4x^3+x^2-4x-6?

Answer 16

Just a moment, please. :)

Answer 17

k

Answer 18

Close. If you distribute 4x across x^2 - 5, that gives you the terms 4x^3 - 20x
Then if you distribute the 1, that gives you x^2 - 5

Answer 19

So the 4x^3 and x^2 terms are correct

Answer 20

oh, so it's 4x^3+x^2-20x-5?

Answer 21

Yes :)

Answer 22

yay thankss

Answer 23

You're welcome!

Answer 24

Got another one?

Answer 25

yup

Answer 26

for f(x)=2x+1 and g(f) = x^2-7, find (f/g)(x)
the answer would be (2x+1)/(x^2-7), x=/= sqrt7 ?

Answer 27

*x=/= +/- sqrt7

Answer 28

Good!

Answer 29

:D

Answer 30

yay

Answer 31

for f(x)=3x+1 and g(x)= x^2-6, find (fog)(4), would i plug in 4 for all x's?

Answer 32

Actually, you can think of these as "nested" functions. (fog)(x) = f(g(x)), so you first find g(4), then take the result and find f of that.

Answer 33

ohh

Answer 34

31?

Answer 35

Let me check :) g(4) = 4^2 - 6 = 16 - 6 = 10, and f(10) = 3(10) + 1 = 31 ... so yes!!!

Answer 36

woooh!

Answer 37

So output of first function g(x), becomes input for f. :)

Answer 38

thanks :P

Answer 39

(3x+1)-(x^2-6) = -x^2+3x-5?

Answer 40

When you distribute the "-" (or -1) across the terms in the second set of parentheses, you have -x^2 + 6 ;)

Answer 41

so it's x^2-3x-7??

Answer 42

Close :) -x^2 +3x + 7

Answer 43

This might clarify it: (3x + 1) - (x^2 - 6) = 3x + 1 -x^2 + 6 = -x^2 + 3x + 7

Answer 44

ohh, i get it. could it be written as 7+3x-x^2?

Answer 45

Yes, certainly. Although generally speaking, polynomials are written with the terms in descending order. So the term with the variable which has the largest exponent would be listed first, and the constant term would be last.

Answer 46

But 7 + 3x - x^2 is certainly an equivalent expression.

Answer 47

i see, thank you :P

Answer 48

i have 2 more, is that ok?

Answer 49

Wlcm :)

Answer 50

Yes

Answer 51

if h(x)=(fog)(x) and h(x)= 3root x+3, find g(x) if f(x)= 3root x+2. i'm completely lost on this one

Answer 52

Please tell me is the "+3" under the radical for h(x)?

Answer 53

Or is it like this:\[\sqrt{3x} + 3\]

Answer 54

it's under the radical

Answer 55

And for f(x), is the 2 under the radical?

Answer 56

yeah

Answer 57

So, h(x) = \[\sqrt{3x + 3}\] , and f(x) = \[\sqrt{3x + 2}\]

Answer 58

yeah

Answer 59

Ok, thanks :)

Answer 60

Thinking for a moment.

Answer 61

k :P

Answer 62

Looking at the expressions for h(x) and f(x), what is different between the two?

Answer 63

+2 and +3

Answer 64

Right! So, basically what you want to do is find an algebraic expression you could plug in for x, in 3x + 2, so that you would get 3x + 3. I can't see a good way of explaining this other than to give the answer and show you, if that is ok?

Answer 65

sure

Answer 66

If we let g(x) be x + 1/3 , then if we put in x + 1/3 for our input in f(x), we would have:

\[\sqrt{3(x+1/3) + 2}\]

Answer 67

Then, when you distribute the 3 and simplify the radicand, what will you have?

Answer 68

3x+3?

Answer 69

Yes :)

Answer 70

So, basically what I was looking for was what expression could we put in for x in f(x), so that we would end up with 3x + 3 under the radical. Does that make sense?

Answer 71

and there would be no radicand left?

Answer 72

No, the square root sign would still be there, but yes, you would have 3x + 3 under the radical.

Answer 73

yeah, that makes sense

Answer 74

oh, ok

Answer 75

So we could say g(x) = x + 1/3

Answer 76

i understand :P

Answer 77

Then f(g(x)) = \[\sqrt{3(x + 1/3) + 2}\] = \[\sqrt{3x + 3} = h(x)\]

Answer 78

yeah i just got that, thanks =]

Answer 79

You're welcome! Any others?

Answer 80

1 more

Answer 81

ok

Answer 82

if f(x)=3x, g(x)=x-3, and h(x)= rootx, find (fogoh)(25)

Answer 83

Let me know what you get. :)

Answer 84

would i begin like 3 (root(x-3)) ? but plug in 25 for x?

Answer 85

You can think of it as f(g(h(x))), so if you start with h(25) and go from there.

Answer 86

h(25) = 5 ... then g(5) = ..

Answer 87

is the answer 6?

Answer 88

Yes :D

Answer 89

But let's look at what you wrote as well, because that is also correct

Answer 90

Ok

Answer 91

You could start like: 3 (rootx - 3) and that should work as well

Answer 92

since root 25 would be 5, then if you subtract 3 that would give 2, and then finally multiply by 2 to get 6.

Answer 93

alright, cool! :D

Answer 94

I mean it would be \[3(\sqrt{x} - 3)\]

Answer 95

ok

Answer 96

Good! :D

Answer 97

those are all the questions i have! thank you so much, i'm ready for my test now :D

Answer 98

Good :) If you would like a little more practice, there are some here at the bottom of the page: http://www.mathsisfun.com/sets/functions-composition.html