OpenStudy (anonymous):

f(x)= x/2+x g(x)= 2/x Find f(g(x))

5 years ago
OpenStudy (diyadiya):

Just replace the x in f(x) with g(x) [that is 2/x]

5 years ago
OpenStudy (diyadiya):

\[\Large f(g(x)) = f( \frac{2}{x}) = \frac{ \frac{2}{x}}{2}+x\]

5 years ago
OpenStudy (diyadiya):

understood? can you continue now?

5 years ago
OpenStudy (anonymous):

Its the computation I am having trouble with.

5 years ago
OpenStudy (anonymous):

wait sorry!! have a question

5 years ago
OpenStudy (diyadiya):

Ok \[\text{You have \to find } \huge f \left( g(x) \right)\] so its given g(x) = 2/x so replace g(x) with 2/x\[\huge f \left( g(x) \right) = f \left( \frac{2}{x} \right)\] is this clear?

5 years ago
OpenStudy (anonymous):

When I replace the terms I get 2/x/2+2/x?

5 years ago
OpenStudy (diyadiya):

Oh Well YES! that was my mistake & i'm extremely sorry about it

5 years ago
OpenStudy (anonymous):

no problem, please continue.

5 years ago
OpenStudy (diyadiya):

\[\large f(x)= \frac{x}{2}+x\]\[\Large f \left( \frac{2}{x} \right) = \frac{\left( \frac{2}{x} \right)}{2}+ \left( \frac{2}{x} \right)\]

5 years ago
OpenStudy (diyadiya):

are you having any trouble in simplifying it?

5 years ago
OpenStudy (anonymous):

yes

5 years ago
OpenStudy (diyadiya):

tell me what is \(\large \frac{\left( \frac{2}{x} \right)}{2}\)

5 years ago
OpenStudy (anonymous):

4x

5 years ago
OpenStudy (anonymous):

4/x

5 years ago
OpenStudy (diyadiya):

i hope you know that 2 can be written as 2/1 so here we are dividing a fraction by another fraction \[ \large \frac{\left( \frac{2}{x} \right)}{ \left( \frac{2}{1} \right)} \]so here we take reciprocal \[ \large \frac{\left( \frac{2}{x} \right)}{ \left( \frac{2}{1} \right)} = \frac{2}{x} \times \frac{1}{2}\] so what do you get now?

5 years ago
OpenStudy (anonymous):

2/2x

5 years ago
OpenStudy (diyadiya):

Yes ,what is 2/2x

5 years ago
OpenStudy (anonymous):

(2/x)/2

5 years ago
OpenStudy (diyadiya):

Noo... 2/2x can be simplified \[\frac{2}{2 \times x} = \frac{\cancel{2}}{\cancel{2} \times x} = ?\](we are cancelling out because 2/2=1)

5 years ago
OpenStudy (anonymous):

1/x

5 years ago
OpenStudy (diyadiya):

Right!!! :D Now what do you have ?

5 years ago
OpenStudy (diyadiya):

\[\Large f \left( \frac{2}{x} \right) = \frac{\left( \frac{2}{x} \right)}{2}+ \left( \frac{2}{x} \right) = \frac{1}{x} + \frac{2}{x}\] Is this clear?

5 years ago
OpenStudy (anonymous):

3/x

5 years ago
OpenStudy (diyadiya):

Very Good! :D

5 years ago
OpenStudy (diyadiya):

i hope you understood!

5 years ago
OpenStudy (anonymous):

yes but my text gives the answer 1/x+1

5 years ago
OpenStudy (diyadiya):

wait.. is your question \[f(x)= \frac{x}{2+x}~~~or~~~ f(x)= \frac{x}{2} + x\]

5 years ago
OpenStudy (anonymous):

The first.

5 years ago
OpenStudy (anonymous):

@Diyadiya

5 years ago
OpenStudy (diyadiya):

i think its the first one, then you'll get the answer 1/x+1 same way we replace g(x) in f(g(x)) by 2/x \[\Large f(\frac{2}{x}) = \frac{\left( \frac{2}{x} \right)}{2+\left( \frac{2}{x} \right)}\] Lets do the denominator first what is 2 + 2/x ?\[2+\frac{2}{x}~~=~~ \frac{2}{1}+ \frac{2}{x}~~=~~\frac{2 \times x}{1 \times x} + \frac{2}{x} ~~= \frac{2x}{x}+ \frac{2}{x}\]**i multiplied by x inorder to have common denominator without which you cannot add two fractions

5 years ago
OpenStudy (anonymous):

yes

5 years ago
OpenStudy (diyadiya):

\[\Large f(\frac{2}{x})= \frac{\left( \frac{2}{x} \right)}{\frac{2x}{x}+ \frac{2}{x}}= \frac{\left( \frac{2}{x} \right)}{\frac{2x+2}{x}}\] Now can you take reciprocal ?

5 years ago
OpenStudy (anonymous):

2/x*x/2x+2= 2/2x+2?

5 years ago
OpenStudy (anonymous):

factor out the 2

5 years ago
OpenStudy (diyadiya):

Yes :D

5 years ago
OpenStudy (anonymous):

:)

5 years ago
OpenStudy (diyadiya):

there you have the answer :D

5 years ago
OpenStudy (anonymous):

thank you

5 years ago
OpenStudy (diyadiya):

Anytime :) sorry i misread the question in the beginning

5 years ago
OpenStudy (anonymous):

no problem your fantastic.

5 years ago
OpenStudy (diyadiya):

Haha Thanks

5 years ago
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