pre calculus (f of g) (x)?

Question

ok having some issues with this problem.


I know the answer is
(1/x)+1

for

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

then for (g of f)(x)

answer is 1/x+1

I am trying to figure out how this answer came to be.
can someone please explain?

Answer 1

Question 1
(f o g) (x) = f ( 1/x ) = (1/x ) + 1

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

Answer 2

F(x) and G(x) might routinely be the integrals of f(x) and g(x). i'm assuming that may no longer what you meant. x^2 + 2x + a million = 3x + 3 x^2 -x - 2 = 0 (x - 2)(x + a million) = 0 x = 2 or -a million

Answer 3

Parenthesis people. 1/(x+1) is not the same as 1/x+1

Answer 4

ok think of the problem this way
g o f (x) = g(f(x))
'g' of 'f' of x.
so you plug in f(x) in place of x in the 'g' function
now f(x) = x+1 ..
and g(x) = 1/x
and now for the g function x = f(x)
therefore
g(x) = 1/f(x) = 1/x+1

Answer 5

plug the function f(x) into the function g(x)
so since
f(x) = x +1
plug that into g(x) wherever you see an x
hence,
1/(x +1)