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given by
g(x) =

1 + 4x

## Let us check this and see what happens.




9 3 1 + 4x
5 3 1 + 4x
+x+
+ 2 = x.
f (g(x)) = (g(x)) + 3g(x) + 2 =
2
2


So far, so good. Now compose the two functions in the opposite order:
p

1 + 4(x2 + 3x + 2)
2

2
3 (2x + 3)
3 4x + 12x + 9
=
= x or
=
2
2

g(f (x)) =

x 3.

We get x for the composite value, as expected, but where did the second answer come
from?
Let us go back to the definition of inverse. In order for a function to have an inverse
from its range to its domain, it must be one-one. This is not the case for the quadratic
f (x), where each value exceeding 1/4 is assumed twice. In order to provide for an inverse,
we restrict the domain of f (x) to a set upon which it is one-one.
Let
f1 (x) = f (x) = x2 + 3x + 2
for x 3/2. The inverse g1 (x) of f1 (x) is the function
g1 (x) =

3 +

1 + 4x

defined for x 1/4. As x increases from 1/4, the value g1 (x) increases from 3/2.
Note that the graphs of f1 (x) and g1 (x) are images of each other reflected in the line y = x.
Now
g1 (f1 (x)) =

3 +

1 + 4(x2 + 3x + 2)
3 +
=
2
1

(2x + 3)2
.
2

Observe that, by definition of the radical, z refers to the positive square root of z. Since
x 3/2, the positive square root of (2x + 3)2 is equal to 2x + 3 and we find that
g1 (f1 (x)) =

3 + (2x 3)
= x.
2

Now define
f2 (x) = f (x) = x2 + 3x + 2
for x 3/2. The inverse of this function is
g2 (x) =

1 + 4x

defined for x 1/4. As x increases from 1/4 the value of g2 (x) decreases from 3/2.
Again, observe that the graphs of f2 (x) and g2 (x) are reflections of each other in the line
y = 3. We have that
g2 (f2 (x)) =

1 + 4(x2 + 3x + 2
3
=
2

p
(2x + 3)2
.
2

This time, since x 3/2, we have that 2x+3 0 and the positive square root of (2x+3)2
is (2x + 3). Therefore
g2 (f2 (x)) =

3 + 2x + 3
3 ((2x + 3))
=
= x.
2
2
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