Functions are here?

Question

Find 2 functions f(x) and g(x) such that
f[g(x)] = x and
g[f(x)] ≠ x

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Answer 1

Answer!
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f(x)=x^2
________
g(x)=±√x
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fog=x
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gof=±√(x)^2 =+/-x
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Hope that this helps! ^_^!!
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Answer 2

You have to do something with the domains of the functions.

For example, let f(x) = tan x on the domain -pi/2 < x < pi/2 and g(x) be its inverse function with the appropriate domain and range. Then since both are injective, f(g(x)) = x and g(f(x)) = x on the domains of g(x) and f(x) respectively, but since the domain of f(x) is a bounded interval, g(f(x)) isn't defined outside this interval. However, since g(x) is defined on all the reals, f(g(x)) = x on all the reals. Thus g(f(x)) ≠ x on the reals, but f(g(x)) = x on the reals.

As a side note, the answer above is incorrect in stating rt(x)^2 = +/-x. If we consider rtx to be a function, then we can only get one output, not two.

Answer 3

some example here:
f(x)= sin(x)
g(x)= cos (x)

then f(g(x)) or you can write

f o g = sin(cos(x))

and its value is not equal to

g o f = cos(sin(x))

lets try, if x=90 , (in deg)
f o g= sin (cos(90))= sin (0) = 0
g o f= cos(sin(90)) = cos (1) ---> is definetely not 0! xD

Answer 4

f(x)=x^2
g(x)=rtx
fog=x
gof=rt(x)^2 =+/-x