Question

how do you gind f(g(n)) if f(n)= 3n-1 and g(n) = n^2 +5?

Answer 1

already solved

Answer 2

how so?

Answer 3

in the function notation f(x) you would replace in the eqaution whatever you put in the grouping symbols.

Suppose f(x)=3x-1
often that is just a number: f(2)=3(2)-1=6-1=5

Answer 4

i understand that if ifs f(2) i would subistute in but with another equation like g(n)= n^2+5 i am confused

Answer 5

You posted this same question earlier

Answer 6

nobody responded to it taco bell man

Answer 7

but you can put anthing inside those grouping symbols, for example

suppose gain f(x)=3x-1
then f(x=2)=3(x+2)-1

does this make sense so far

Answer 8

well with this problem currently i understand to this point \[f(n^2+5)=\]

Answer 9

after what i have is where i am lsot because i am unsure what to do next

Answer 10

in your case, you have f(n)=3n-1 and g(n)=n^2+5

if you want to find f(g(n)) just put g(n) in for the variable in function f for example\[f \left( g(n) \right)=f \left( x^2+5 \right)=3\left( x^2+5 \right)-1\]

Answer 11

all you have to do from here is distribute the 3 and simplify\[=3x^2+15-1=3x^2+14\]

Answer 12

sorry i accidentally changed n to x

Answer 13

its fine and i think i understand it better now but my next problem is confusing me more i think, can you help?

Answer 14

please?

Answer 15

yes

Answer 16

then the next one is g(n)=n+4 and h(n)= 4n +5 but what is confusing me is this segment here. find g(h(10))

Answer 17

right here is the link to your previous question.
http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e7296ed0b8b247045ca52be

Answer 18

ooops sorry taco bell but i didnt get a notifaction that it was answered

Answer 19

The 10 just means evaluate the function that you find at 10. Let's do it one step at a time.

Answer 20

haha it is no big deal, i was just letting you know and it has a pretty good detailed explanation

Answer 21

so in order to work the 10 i must do h(n)=4(10) +5 ? then mixed this sum into the other equation?

Answer 22

First just find g(h(n))=g(4n+5)
\[g(h(n))=\left( 4n+5 \right)+4=4n+9\]now you can find g(h(10))\[g(h(10))=4(10)+9=49\]

Answer 23

so i must substite the problem first in the the other equation before putting the 10 in ?

Answer 24

no, but is one way of doing it. you could sub the 10 into h first then sub that number, 45, into g

Answer 25

is the way you showed me the easiest way?

Answer 26

ok i am understanding it way more better then i did 30 minutes ago, i am going to list another problem and do it step by step, i would like you to help me by seeing if i am doing it correctly

Answer 27

if you are going to evaluate the function at more than one domainn value, you would do it the way I showed you; if you are just going to do it once, the second way might be better.

i chose the way i showed you because i thought you wanted the practive in finding function of the form f(h(n))

Answer 28

ok

Answer 29

f(x) =2x
g(x)= 4x-1
find f(g(-9))

Answer 30

i first would do f(4x-1)=2(4x-1)
then f(4x-1)= 8x-2

Answer 31

then the final step would be f(g(-9))= 8(-9)-1 which equals -74 so therefore
f(g(-9)) = -74

Answer 32

is that correct?

Answer 33

yes, correct!

Answer 34

the other way would be find g(-9) first:
g(-9)=4(-9)-1=-36-1=-37
then find f(g(-9))=f(-37)=2(-37)=-74

Answer 35

i also see it that way also, now i know 2 ways to figure it out, much better for me

Answer 36

well what do i do if the symbol is squared like this

Answer 37

g(t)= t^2 =1
h(t)= t-4

Answer 38

and what are you trying to find?

Answer 39

g(h(4))

Answer 40

i think this is only problem i need then i can undertstand all problems that wre close to the ones i discussed

Answer 41

can I assume g(t)=t^2-1 and not t^2=1

Answer 42

oops i meant g(t)=t^2+1

Answer 43

what is confusing me is the squared, i am unsure of what to do hence the problem i cant think i can do it

Answer 44

Method 1:
h(4)=4-4=0
then
g(h(4))=g(0)-0^2+1=0+1=1

Method 2:
g(h(t))=g(t-4)=(t-4)^2+1=t^2-8t+16+1=t^2-8t+17
then
g(h(4))=4^2-8(4)-15=16-32+17=1

Answer 45

remember\[(a-b)^2=a^2-2ab+b^2\]

Answer 46

but where do you get the 8t from?

Answer 47

in method 2

Answer 48

using the formula i showed t=right after method 2 where a=t and b=4\[(t-4)^2=t^2-2(t)(4)+4^2\]

Answer 49

-2*t*4=-8t

Answer 50

the 2nd method confuses me but method 1 i totally understand

Answer 51

you can always just use method 1

Answer 52

i believe i understand the method now, now i wanna ask another since it has another variable but i feel that i am abusing your help

Answer 53

i do this to help out

Answer 54

then last question i believe i would ask is
f(a)=a-3
g9a)=-a-1
find f(g(-2a))

Answer 55

acutlly now that i look at it, it fairly easy

Answer 56

you do it, i'll check your work then

Answer 57

i think what i do is g(a)=-2a-1

Answer 58

then -2a-1-3 would be-2a-4 so there fore f(g(-2a))= -2a-4

Answer 59

you get the mando seal of approval!

Answer 60

yay but wait i forgot its g(a)=-a^3

Answer 61

but wait,

Answer 62

the 2 is a negative and -a^3 is a negative so it will be 2a^3 so it will be 8a

Answer 63

so then will be the correct answer be f(g(-2a))= 8a-4?

Answer 64

wait, what is function f

Answer 65

funtcion f is f(a)=a-3

Answer 66

i do think you made a sign error earlier g(-2a)=-(-2a)-1=2a-1
then f(g(-2a))=f(2a-1)=2a-1-3=2a-4

Answer 67

yea i put f(g(-2a))=-2a-1 but i mistanekly wrote the g function wrong its acutllay
g(a)=-a^3-1. i forgot the third power, but i know the sign will be positive and will that make 8a?

Answer 68

?

Answer 69

g(-2a)=8a^3-1
f(g(-2a))=8a^3-1-3=8a^3-4

Answer 70

wait its 2a^3-3-1, what do i do with the third? do i do 2^3 which is 8 and add the a or just stays 2a^3?

Answer 71

-(-2a)^3=-(-2)^3a^3=-(-8)a^3=8a^3

Answer 72

both factors the -2 and the a have to be cubed

Answer 73

ohhhh ok thanks, going to be going now, YOUR A AMAZING PERSOn, thanks for the help and KEEP IT UP

Answer 74

later gtg myself