Find the inverse
of y
= x^{2} + 1, x>
0, and determine
whether the inverse is a function.

You'll
notice that the only difference between this and the previous
example
is that the domain has been restricted to the positive
x-axis
this time. Here's the graph:

Since
I already figured out the domain and range, I know that I
have to choose the positive square root:

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

Here's
the graph:

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

If you've studied function
notation, you may
be starting with "f(x)"
instead of "y".
In that case, start the inversion process by renaming f(x) as
"y";
find the inverse, and rename the resulting "y" as "f^{–1}(x)".
It's usually easier to work with "y". Warning: This notation
is misleading; the "minus one" power in the function notation
means "the inverse function", not "the reciprocal of".
Don't confuse the two.

Find the inverse
of y
= ^{–2} / _{(x – 5)},
and determine whether the inverse is also a function.

Since the variable is
in the denominator, this is a rational function. Here's the algebra:

The
original function:

I
multiply the denominator up to the left-hand side of the equation:

I
take the y
through the parentheses:

I
get the x-stuff
by itself on one side of the "equals" sign:

Then
I solve for x:

And
then switch the x's
and y's:

This is just another
rational function. The
inverse function is y
= ^{(5x – 2)} / _{x}

Find the inverse
of f(x)
= –sqrt(x – 2), x>
2. Determine whether
the inverse is also a function, and find the domain and range of the
inverse.

The
domain restriction comes from the fact that x
is inside a square
root. Usually I wouldn't bother writing down "x> 2",
because I know that x-values
less than 2
would give me negatives
inside the square root. But the restriction is useful in this
case because, together with the graph, it will help me determine
the domain and range on the inverse:

The domain is
x>
2; the range (from
the graph) is y< 0. Then
the domain of the inverse will be x< 0; the
range will be y> 2. Here's
the algebra:

The
original function:

Rename
"f(x)"
as "y":

Solve
for "x
=":

Switch
x
and y:

Rename
"y"
as "f-inverse".
Since I already figured out the domain and range, I know which
half of the quadratic I have to choose:

Then the
inverse y
= x^{2} + 2 is
a function, with domain x<
0 and range
y>
2.