How do i simplify g(f(x)) and g(g(x))?

Question

f(x) = x+(1/x)

g(x) = (x+1)/(x+2)

I need g(f(x)) and g(g(x)). How would i get the simplest form of that?

Answer 1

basically what you do for this is just put your F(x) equation into all your X vales oif your G(x) equation.

so what you have is   ((x+(1/x))+1)/((x+(1/x))+2)

now you try the next one on your own.  *hint* just put g(x) into every x value for your equation of g(x)

Answer 2

I beg your pardon. In english lol. I don't understand any of that really sorry.

Answer 3

for g(f(x)) plus in f(x) for x every time in g(x)

g(x)=(x+1)/(x+2)

g(f(x))=(f(x)+1)/(f(x)+2)

g(f(x))=(x+(1/x)+1)/(x+(1/x)+2)

g(g(x)) is the same except plug g(x) in for x

g(x)=(x+1)/(x+2)

g(g(x))=(g(x)+1)/(g(x)+2)

g(g(x))=((x+1)/(x+2)+1)/((x+1)/(x+2)+2...

Answer 4

??????

Answer 5

The question is simply asking you that in g(x), substitute "x" with all of the values of function x; (f(x)) and then simplfy the new function.the final ans. shoud be;

g(f(x)) =(x^2+x+1)/(x^2+x+2).

good luck.

Answer 6

oh ****.

haha; how are we supposed to know that? haha

Answer 7

f(x) = x + (1/x)

g(x) = (x + 1) / (x + 2)

g(f(x))

= g[x + (1/x)]

= [x + (1/x) + 1] / [x + (1/x) + 2]

= [(x^2/x) + (1/x) + (x/x)] / [(x^2/x) + (1/x) + (2x/x)]

= [(x^2 + 1 + x) / x] / [(x^2 + 1 + 2x) / x]

= (x^2 + x + 1) / (x^2 + 2x + 1)

g(g(x))

= g[(x + 1) / (x + 2)]

= {[(x + 1) / (x + 2)] + 1} / {[(x + 1) / (x + 2)] + 2}

= {[(x + 1) / (x + 2)] + [(x + 2)/(x + 2)]}

/ {[(x + 1) / (x + 2)] + [2(x + 2) / (x + 2)]}

= {[(x + 1) + (x + 2)] / (x + 2)] / {[(x + 1) + 2(x + 2)] / (x + 2)}

= [(x + 1) + (x + 2)] / [(x + 1) + 2(x + 2)]

= (2x + 3) / (3x + 5)