f°g(x)=f(g(x))
This is called the composition
of the functions f and g
and the idea behind it is
to do the two operations
of the function one after
another. If you think of
functions as little machines
it is like linking up the
two machines together to
form a new machine. This
goes in reverse order of
the way it is written, because
this makes the order come
out right with the nested
parentheses. What is happening
here is that the x goes
into the g machine first
and out comes g(x). Then
the g(x) goes into the f
machine and out comes f(g(x)).
So for example, if f(x)=x+1
and g(x)=x
2 ,
then since g is the function
that squares a number, and
f is the function that adds
1 to a number, the composition
function f
og would be the function that first squares the number and then
adds 1 to the result of
that, or f
og(x)=x
2+1.
For more complicated functions
it might be difficult to
keep track of it that way,
but that is not a problem,
because if you just follow
the definition, the notation
is designed to make it come
out right without thinking
about it too much. So, for
example, in this problem
you could also figure it
out simply by writing down
fog(x)=f(g(x))
and replacing things
with what they are equal
to.
fog(x)=f(g(x))=f(x2)
At this point you have
to make sure to use function
notation correctly (See
my article Using Function Notation.) If it helps, rewrite
the definition for f without
the x. f(
)=(
)+1 and fill in the holes.
(I will be doing that
in all of the examples
below, but you don't need
to do it if you can keep
track of things simply
by replacing the x with
what is in the parentheses.)
f(x2)=(x2)+1=x2+1.
In a similar way, we
can find g°f.
This should be the function
that adds 1 and then squares.
gof(x)=g(f(x))=g(x+1).
g(
)=(
)2
so filling in the x+1,
we get
g(x+1)=(x+1)2
One interesting thing
to notice from this is
that composition of functions
is not a commutative operation.
fog
is not the same as gof.
Often to get practice
with the idea of composition
of functions and its notation
students are asked to
find f°g
and gof,
sometimes called fog and
gof. When you do that
you should be suspicious
if you get the same answer
for both of them, because
most of the time that
won't be right. In the
following examples the
instruction is to find
fog
and gof.
Example 1:
Solution:
fog(x)=f(g(x))=
Writing the definition of
f without the x's, we get
f(
)=(
)3.
Then fill in the holes.
gof(x)=g(f(x))=g(x3)
Writing the definition
of g without the x's,
we get
Then fill in the holes.
Example 2:
Solution:
fog(x)=f(g(x))=f(x2+x+2)
Writing the definition
of f without the x's we
get
Then fill in the holes.
Writing the definition of
g without the x's we get
g(
)=(
)2+(
)+2 .
Then fill in the holes
and simplify.
Example 3:
Solution:
Writing the definition of
f without the x's we get
Then fill in the holes and
simplify.
Here in order to simplify
the complex fraction I am
multiplying top and bottom
by the least common multiple
of all of the denominators,
which is x.
Writing the definition
of g without the x's we
get
Then fill in the holes
and simplify.
Here I am inverting and
multiplying, because that
is how you divide fractions.
Example 4:
f(x)=4, g(x)=x2+3
Solution:
fog(x)=f(g(x))=f(x2+3)
This one can be a little
confusing, so we have
to be careful. Writing
down the definition of
f without the x's we get
f(
)=4.
Then filling in the holes,
there is no hole on the
right side, so
f(x2+3)=4,
Since there is nothing
to fill in on the right
side, that's all there
is to do.
gof(x)=g(f(x))=g(4)=42+3=16+3=19
This time I am filling
in the hole with a number,
because no matter what
you put into the g function
you get 4 out of it.
g( )=(
)2+3
Fill in the hole here
with a 4.
Example 5:
f(x)=4, g(x)=3
Solution:
fog(x)=f(g(x))=f(3)=4
Basically the idea here
is that f(x)=4 means that
f(anything)=4, including
3, or if as before we
write the definition of
f without the x's we get
f( )=4
If we fill in the hole
with a 3 we get, just
as above, f(3)=4, because
there is no hole to fill
in on the right side.
gof(x)=g(f(x))=g(4)=3
g(
)=3
Fill in the blank with
a 4 to get it.
