I presume this is functional composition, not function multiplication. The open circle normally means functional composition.
The output of f(x) becomes the input to h(x) because f is closer to x than h is.
(h°f)(x) = h(f(x)) = h(-3x) = 1/[(-3x)+1]
We must exclude all values in the domain of f(x) which cannot be evaluated. Since f(x) = -3x, there are no values which cannot be evaluated.
If the denominator of 1/(x+1) is zero, the function is undefined. x +1 = 0 means that x = -1.-3x = -1 when x = 1/3. So that is the only value that has to be excluded from the domain of hºg. In interval notation,
The domain of hºf = ( -∞ , 1/3 ) ∪ ( 1/3 , ∞ ).
