Answer: [tex]g^{-1}(x) = \frac{1-4x}{x}[/tex]
Domain of the inverse is any real number but x cannot equal zero
Domain in set builder notation = [tex]\{x | x\in\mathbb{R} , \ x \ne 0\}[/tex]
Domain in interval notation = [tex](-\infty, 0) \cup (0, \infty)[/tex]
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Let's make h(x) be the inverse of g(x)
g(x) = 1/(x+4) is the same as y = 1/(x+4).
To find the inverse, we swap x and y, then solve for y like so....
y = 1/(x+4)
x = 1/(y+4) .... x and y swap
x(y+4) = 1
xy+4x = 1
xy = 1-4x
y = (1-4x)/x
h(x) = (1-4x)/x is the inverse of g(x)
We can verify this by showing that g(h(x)) = x and h(g(x)) = x. I'll leave that for you to confirm.
The domain of h(x) is the set of real numbers x such that x cannot be 0. So x can be anything but 0. In set builder notation, we would say [tex]\{x | x\in\mathbb{R} , \ x \ne 0\}[/tex] and in interval notation that is [tex](-\infty, 0) \cup (0, \infty)[/tex]
We kick out x = 0 so that (1-4x)/x doesn't have any division by zero errors.
