Question

"given f(x)=x^2 what is the domain of g(x)=f(x+2)-5"

Answer 1

Any kind of transformation on a parabola will not change its domain.

Answer 2

this is a composition isn't it?

Answer 3

Having said that, g(x) domain would be same as f(x)

Answer 4

It is a composition, you can look at it as transformation also.

Answer 5

Oops ! I was wrong

Answer 6

We need to work this as composition only

Answer 7

so it wouldn't just be the domain of: g(f(x))=(x^2 +2)-5

Answer 8

is that the proper composition? and then you find the domain that way?

Answer 9

There is a trick, which simplifies finding the domain

Answer 10

g(x)=f(x+2)-5

Answer 11

what values we feeding in to g(x) ? Those all are output given by f only

Answer 12

so x is in the domain of real numbers with no restrictions? like x^2?

Answer 13

g(x) never gets a value less than 2 to its input.

Answer 14

Domain of g(x) : x > = 2

Answer 15

why???

Answer 16

um this is kind of confusing. would you mind just telling me if I composed the function properly?

Answer 17

g(f(x))=(x^2 +2)-5

Answer 18

hold on the composition for a bit.

Domain of f(x) : all reals
Range of f(x) : x > 0

Domain of g(x) is the Range of f(x)

Answer 19

On second thoughts, Domain of g(x) is : x >= 0, which is the range of f(x)

Answer 20

Still its confusing ?

Answer 21

g(x) = (x + 2)^2 - 5
= x^2 + 4x - 1

the domain of f(x) is all real x
the domain of g(x) is the same as f(x)

Answer 22

thats wrong cwrw, we need to see f(x) and g(x) as composition. your working makes g(x) isolated

Answer 23

the range of g(x) is [-5, + infinity)

Answer 24

g(x) never gets a negative value at its input because, all values it gets in input are fed from the output of f

Answer 25

yes i see your point - but the question is somewhat ambiguous

Answer 26

i'm confused also lol

Answer 27

^agree it seems ambiguous first read

Answer 28

I twisted the answer so many times before seeing it a bit clearly.. seems Cutie ran away lol

Answer 29

yea

Answer 30

can g(x) be a composition?

if so whats the function h in g(x) = h(f(x) ?

Answer 31

i would see it like this proly :

h is a simply composition

g is a mix of composition and transformation

Answer 32

*simple

Answer 33

sorry, i was really just confused about the wording too. does the way the f was placed mean the question is really what is the domain of g(fx) or something else?

Answer 34

we can see it like this :-

let h(x) = x+2
g(x) = f(h(x)) - 5

Answer 35

we still need to find domain of g(x) only

Answer 36

if you think f(h(x)) as variable t

g(x) = t-5

Answer 37

t can take only values : t >= 0

Answer 38

so domain is [0, +inf)

Answer 39

i've just input the problem into the wolfram alpha math software program.
it ignored the f\9x) = x^2 bit and came back with the domain of g(x) being all real numbers

i think theres something wrong with the question

Answer 40

interesting, can you give the link

Answer 41

The domain is all real numbers, g(x) is defined for all inputs x in R.

Answer 42

http://www.wolframalpha.com/input/?i=given+f%28x%29%3Dx%5E2+what+is+the+domain+of+g%28x%29%3Df%28x%2B2%29-5

Answer 43

question clearly says g is a composition of f

Answer 44

hhmmm - i dont think so

Answer 45

Even if it is a composition, the domain is still all R. However if g(x)=Log(x) it would be another story,

Answer 46

@cwrw238 see this

http://www.wolframalpha.com/input/?i=given+f%28x%29%3Dlog%28x%29++what+is+the+range+of+g%28x%29%3Df%28x%2B2%29-5

Answer 47

would like to see wat @UnkleRhaukus thinks

Answer 48

by definition
\[f(x+2) = (x+2)^{2}\]
thus
\[g(x) = (x+2)^{2} -5\]
so domain of g is all real numbers
@rsadhvika had it right at first then a lot of over analyzing i guess

Answer 49

f(x) is a polynomial function whose domain is all reals: