x ≠ -2 is true for f(x), but we are interested in f[g(x)]. In that case, the restriction is _g(x)_ ≠ -2. Solve that to find the restriction in terms of x. 4/(x - 1) ≠ -2 -2(x - 1) ≠ 4 x - 1 ≠ -2 x ≠ -1 This is the same restriction that already applies to f[g(x)] by its own definition, so an additional note is not necessary.
I have some worry if we consider fog as a one function h definition Domaine is R-{-1} h(x) = (x-1)/2x+2 but f(g(x)) =f(4/x-1) is defined when g is defined and f is defined. But fog or h is defined because x-1 come to the top of fraction.
honestly is a little bit confusin I asked him around it. the combination have an image and g is undefined in the point and f is continue and have a limite in infinity.
