The customary method of
finding the inverse is some variant of the method I'm going to use below.
Whatever method you use, make sure you do the exact same steps in the
exact same order every time, so you remember those steps when you get
to the test.

Find the inverse
of y
= 3x – 2.

Here's how the process
works:

Here's
my original function:

Now
I'll try to solve for "x
=":

Once
I have "x
=",
I'll switch x
and y;
the "y
=" is the
inverse.

Then the
inverse is y
= ^{(x + 2)} / _{3}

If you need to find the
domain and range,
look at the original function and its graph. The domain of the original
function is the set of all allowable x-values;
in this case, the function was a simple polynomial, so the domain was
"all real numbers". The range of the original function is all
the y-values
you'll pass on the graph; in this case, the straight line goes on for
ever in either direction, so the range is also "all real numbers".
To find the domain and range of the inverse, just swap the domain and
range from the original function.

Find the inverse
function of y
= x^{2} + 1,
if it exists.

There will be times when
they give you functions that don't have inverses.

From the graph,
it's easy to see that this function can't possibly have an inverse,
since it violates the Horizontal Line Test:

What will this look like
when I try to find the inverse algebraically? The Vertical
Line Test says
that I can't have two y's
that share an x-value.
That is, each x
has to have a UNIQUE corresponding
y
value. But look at what happens when I try to solve for "x
=":

My
original function:

Solving
for "x
=":

Well, I solved for "x
=",
but I didn't get a UNIQUE "x
=".
Instead, I've shown that any given x-value
will actually correspond to two different y-values,
one from the "plus" on the square root and the other from
the "minus".

The inverse is not
a function.

Any time you come up with
a "±" sign, you can be pretty sure that the inverse isn't a
function.

The only difference
between this function and the previous one is that the domain
has been restricted to only the negative half of the
x-axis.
This restriction makes the graph look like this:

This function will
have an inverse that
is also a function. Just about any time they give you a problem where
they've taken the trouble to restrict the domain, you should take care
with the algebra and draw a nice picture, because the inverse probably
is a function, but it will probably take some extra effort to show this.
In this case, since the domain is x< 0 and
the range (from the graph) is 1
<y,
then the inverse will have a domain of 1
<x and
a range of y< 0. Here's
how the algebra looks:

The
original function:

Solve
for "x
=":

By
figuring out the domain and range of the inverse, I know that
I should choose the negative sign for the square root:

Now
I'll switch the x
and y;
the new "y
=" is the
inverse:

(The "x>
1" restriction
comes from the fact that x
is inside a square root.)

So the
inverse is y
= –sqrt(x – 1), x> 1,
and this inverse is also a function.