Example 1 Given f (x) = 2 + 3x
− x2
and g (x) = 2x − 1 evaluate each of the following.
(a) ( f + g )(4)
(b) g − f
(c) ( fg )(x)
By evaluate we mean one of two things depending on what
is in the parenthesis. If there is a number in the parenthesis then we want a
number. If there is an x (or no parenthesis, since that implies and x)
then we will perform the operation and simplify as much as possible.
(a) ( f + g )(4)
In this case we’ve got a number so we need to do some
function evaluation.
( f + g ) = f (4) + g
(4)
= ( 2 + 3 (4) – 42 ) + (2 (4) – 1)
= – 2 + 7
= 5
(b) g − f
Here we don’t have an x or a number so this
implies the same thing as if there were an x in parenthesis. Therefore,
we’ll subtract the two functions and simplify. Note as well that this is written
in the opposite order from the definitions above, but it works the same way.
g
– f = g (x) – f (x)
= 2x – 1 – (2 + 3x
– x2)
= 2x – 1 – 2 – 3x +
x2
= x2 – x
– 3
(c) ( fg )( x)
As with the last part this has an x in the
parenthesis so we’ll multiply and then simplify.
( fg ) (x) = f (x)
g (x)
= (2 + 3x – x2)(2x – 1)
= 4x + 6x2 – 2x2 – 2 – 3x
+ x2
= – 2x2 + 7x2
+ x – 2
(d) f (0)
In
this final part we’ve got a number so we’ll once again be doing function
evaluation.
Example 2 Given f (x) = 2 + 3x
− x2 and
g (x) = 2x − 1 evaluate each of the following.
(a)
( fg )(x)
(b)
( f ∘ g )(x)
(c)
(g ∘ f )(x)
Solution
(a)
These are the same functions that we used in the first set of
examples and we’ve already done this part there so we won’t redo all the work
here. It is here only here to prove the point that function composition is NOT
function multiplication.
Here is the multiplication of these two functions.
( fg )(x) = −2x3 + 7x2 + x – 2
(b)
Now, for function composition all you need to remember is that we
are going to plug the second function listed into the first function listed. If
you can remember that you should always be able to write down the basic formula
for the composition.
Here is this function composition.
(f ∘ g
)(x) = f [ g (x) ]
= f [2x
– 1]
Now, notice that since we’ve got a formula for g (x) we
went ahead and plugged that in first.
Also, we did this kind of function evaluation in the
first section we
looked at for functions. At the time it probably didn’t seem all that useful to
be looking at that kind of function evaluation, yet here it is.
Let’s finish this problem out.
(f ∘ g
)(x) = f [ g (x) ]
= f [2x
– 1]
= 2 + 3(2x – 1) – (2x – 1)2
= 2 + 6x – 3 – (4x2
– 4x + 1)
= – 1 + 6x – 4x2 +
4x – 1
= – 4x2 + 10x – 2
Notice that this is very
different from the multiplication! Remember that function composition is NOT
function multiplication.
(c)
We’ll not put in the detail in this
part as it works essentially the same as the previous part.
(g ∘ f
)(x) = g[ f (x) ]
= g [2 +
3x – x2]
= 2 (2 + 3x – x2) – 1
= 4 + 6x – 2x2 – 12
= – 2x2 + 6x + 1
Notice that this is NOT
the same answer as that from the second part. In most cases the order in which
we do the function composition will give different answers.
Example 3 Given f (x) = x2 − 3 and h(x) = x + 1 evaluate each of the following.
(a)
( f ∘ h)(x)
(b)
(h ∘ f )(x)
(c)
( f ∘ f )(x)
(d)
(h ∘ h)(8)
(e)
( f ∘ h)(4)
Solution
(a)
( f ∘ h)(x)
Not
much to do here other than run through the formula.
= x + 1 – 3
= x – 2
(b)
(h ∘ f )(x)
Again, not much to do
here.
(h ∘ f )(x) = h [ f (x) ]
= h [ x2 – 3 ]
(c)
( f ∘ f )(x)
Now in this case do not
get excited about the fact that the two functions here are the same.
Composition works the
same way.
(f ∘ f )(x) = f [ f (x) ]
= f [ x2 – 3 ]
= (x2
– 3)2 – 3
= x4 – 6x2
+ 9 – 3
= x4 – 6x2
+ 6
(d)
(h ∘ h)(8)
In this case, unlike all
the previous examples, we’ve got a number in the parenthesis instead of an x,
but it works in exactly the same manner.
(h ∘ h)(8) = h [ h (8) ]
(e)
( f ∘ h)(4)
Again, we’ve got a number here. This time there are actually two
ways to do this evaluation.
The first is to simply use the results from the first part since
that is a formula for the general function composition.
If we do the problem that way we get,
( f ∘ h)(4) = 4 − 2 = 2
We could also do the evaluation directly as we did in the previous
part. The answers should be the same regardless of how we get them. So, to get
another example down of this kind of evaluation let’s also do the evaluation
directly.
(f ∘ h)(4) = f [ h (4) ]
So, sure enough
we got the same answer, although it did take more work to get it.
Example 4 Given f (x) = 3x − 2 and evaluate each of the following.
(a)
( f ∘ g )(x)
(b)
(g ∘ f )(x)
Solution
(a)
Hopefully, by this point
these aren’t too bad.
( f ∘ g)(x) = f [ g (x) ]
Looks
like things simplified down considerable here.
(b)
All we need to do here is use the formula so
let’s do that.
( g ∘ f)(x) = g [ f (x) ]
= g [3x
– 2]
So, in this
case we get the same answer regardless of the order we did the composition in.
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