Question

What is g^-1(x)? What is the domain of g^-1(x)? Please help!! It’ll be greatly appreciated!!

Answer 1

Answer:  $g^{-1}(x) = \frac{1-4x}{x}$

Domain of the inverse is any real number but x cannot equal zero

Domain in set builder notation = $\{x | x\in\mathbb{R} , \ x \ne 0\}$

Domain in interval notation = $(-\infty, 0) \cup (0, \infty)$

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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 $\{x | x\in\mathbb{R} , \ x \ne 0\}$ and in interval notation that is $(-\infty, 0) \cup (0, \infty)$

We kick out x = 0 so that  (1-4x)/x doesn't have any division by zero errors.