Question

Verify that f and g are inverse functions mathematically :
f(x)=(3-x)/ x g(x)=(3)/x+1

solve: f(g(x))

Answer 1

u have
\[ \large f(x)=\frac{3-x}{x}\qquad g(x)=\frac{3}{x+1} \]
right?

Answer 2

yes.

Answer 3

to verify they r inverses of each other just do this
\[ \large (f\circ g)(x)=f(g(x)) \]
and
\[ \large (g\circ f)(x)=g(f(x)) \]

Answer 4

f(g(x))=f((3)/x+1)=(3-(3)/x+1)/ ( (3)/x+1)= x+1-1 =x

Answer 5

u should get \(x\) when finished

Answer 6

Yea thats what i did but for f(g(x)) i somehow got 1so im going to redo it but i did get X for g(f(x))

Answer 7

let's do it then

Answer 8

\[ \large (f\circ g)(x)=f(g(x))=f\left(\frac{3}{x+1}\right)= \]
\[ \large =\frac{3-\frac{3}{x+1}}{\frac{3}{x+1}}=
\frac{\frac{3(x+1)-3}{x+1}}{\frac{3}{x+1}} \]

Answer 9

ok so far?

Answer 10

hmm how did you get the numerator to equal that? I thought you would have to rationalize 3 to have a denominator of x+1

Answer 11

u do understand the substitution, though. right?

Answer 12

Yes I got that.

Answer 13

now in the numerator i just added the fractions:
\[ \large a+\frac{b}{c}=\frac{ac+b}{c} \]

Answer 14

so you multiplyed -1 * x+1 then added 3?

Answer 15

i don't understand your question.

Answer 16

i multiplied 3 times x+1 and -1 times 3

Answer 17

well after simplifying
\[ \large =\frac{3x+3-3}{3}=\frac{3x}{3}=x \]

Answer 18

oh ok I see sorry i had put the input for x in parenthesis which made it look like 3-1(3/x+1) and ok yea i got that now thanks!

Answer 19

u r welcome