How do you find domain and range of inverse functions f(x) = 1/(x-2)f(x)=1x−2?

2 Answers
MeneerNask · Jim H
Jul 27, 2015

Most important limitation is that the denominator may not be 00

Explanation:

So the domain is limited by x!=2x≠2

As x->2x→2 the fraction gets larger and larger, whether you go to 22 from above or below. In "the language":

lim_(x->2^-) f(x)=-oo and lim_(x->2^+) f(x)=+oo

If x gets larger the fraction will become smaller, but never quite reaching 0, so the range is f(x)!=0 or:

lim_(x->-oo) f(x)=0 and lim_(x->+oo) f(x)=0
graph{1/(x-2 [-10, 10, -5, 5]}

x=2andf(x)=0 are called asymptotes .

Jim H
Jul 27, 2015

The domain of an inverse function is the range of the function and vice versa.

Explanation:

For f(x) = 1/(x-2), the domain of f is all reals except 2, so the range of f^-1 is all reals except 2.

The range of f is all reals except 0, so the domain of f^-1 is all reals except 0.

Notice that is we solve y = 1/(x-2) for x, we get:

y(x-2)=1

xy-2y=1

xy=2y+1

x = (2y+1)/y

We can see from this that for the original function, f, we can get every number for y except 0.
That is the range of f and the domain of f^-1

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