Question

f(x)=2x+1 and g(x)=x^(2)-x

Calculate fog and gof and domains of the functions.

Answer 1

help here.

Answer 2

\[f(x)=2x+1\]
\[g(x)=x^2-x\]
\[f\circ g(x)=f(g(x))=f(x^2-x)=2(x^2-x)+1\] then maybe algebra

Answer 3

\[f\circ g(x)=2x^2-2x+1\]
domain all real numbers because you are working with polynomials

Answer 4

yes ok . I have solved the composite function. problem for domain. how to find domain?

Answer 5

ok all real numbers except inf. right?

Answer 6

step one: get rid of circle notation
\[g\circ f(x)=g(f(x))\]
step two: replace the general f by the specific one you have
\[g(f(x))=g(2x+1)\] step three: evaluate the function at your new input, that is replace "x" by whatever is in the parentheses
\[g(2x+1)=(2x+1)^2-(2x+1)\] step four: clean up with algebra

Answer 7

all real numbers is all real numbers
infinity not a number

Answer 8

i know how to solve composite function.
someone said me domain is located as individual functions not combined functions.
for example.
f(x)=2sqrt(x-1) and g(x)=sqrt(x-1)
and want to calculate f(x)/g(x)
so
f(x)/g(x) = 2sqrt(x-1)/sqrt(x-1)
so here we will look at denominator for domain. although function is not fully computed.
if i complete my calculation then result would be
f(x)/g(x)=2
so here we can't estimate domain.

But in my given question. First we should solve the composite function and then find domain, not like above example. right?

Answer 9

no you are right, you need the domain of the "inside" function first
i ignored that because you are working with polynomials so domain is all real numbers for each, and also for the composite

Answer 10

f(g(x))=2(x^2-x)+1

you mean at this stage we have to locate domain. right?

Answer 11

for example if
\[f(x)=x^2\] and
\[g(x)=\sqrt{x-1}\] then
\[f\circ g(x)=f(g(x))=f(\sqrt{x-1})=\sqrt{x-1}^2=x-1\] but the domain is determined by the domain of
\[g(x)\] which is
\[[1,\infty)\]

Answer 12

OK dude. I got you. for composite functions domain will always defined by taking inner function.
Thanks dude.