Transcript Combining and inverting functions

Combining and inverting
functions
Function notation
f ( x)  x2  6x  4
This notation is technically wrong since f ( x) should really be
reserved for the output of the function f when the input is x,
not for the function f itself.
f :x
x2  6 x  4
Domain and range
If we write
f :x
x 1
without specifying the domain we should assume that f has the largest
domain possible i.e. x  1.
We say that f has natural domain of x  1 and should think of the full
function definition as
f :x
x  1, x  1
Note that
f :x
x  1, x  1 and f : x
are different functions.
x  1, x  2
Note, for example, that sin x, x  has range  1  y  1 whereas
sin x, 0  x  90 has range 0  y  1.
Function composition
To form the composite function g f , the domain D of f must be chosen so
that the whole of the range of f is included in the domain of g. The function g f
is then defined as g f : x
Example : The function x
function g f , where f : x
g  f ( x)  , x  D.
x( x  3) can be considered as the composed
x( x  3) and g : x
x.
The natural domain of g is x  , x  0, so we require the range of f to be
included in this set. The solution of the inequality x( x  3)  0 is x  0 or x  3.
So the natural domain of g f is x  , x  0 or x  3.
With this domain, the range of g is y  , y  0 which is therefore the range of
the composed function g f .
Do Exercise 11A, pp.162-164
Inverse functions
A function f defined for some domain D is one - one if, for each number
y in the range of f , there is only one number x  D such that y  f ( x).
The function with domain R defined by f 1 : y
the inverse function of f .
x, where y  f ( x), is
To consider geometrically whether a function is one-one we use the horizontal
line test. To prove algebraically a function f is one-one we show that
f ( x1 )  f ( x2 )  x1  x2 for all x1 , x2 in the domain of f .
Both f 1 f ( x)  x and f f 1 ( y)  y are called identity functions but may have
different domains.
Do Exercise 11B, pp. 169-172
Inverse functions
x 2  2 x so that an inverse function exists.
Example : Restrict the domain of f : x
Find an expression for f 1.
One possible domain restriction is x  , x  1,
which has accompanying range y  , y  1.
If y  x 2  2 x, then x 2  2 x  y  0 is a quadratic
in x, with roots x 
2  4  4y
 1 1 y.
2
Since x  1, we choose the positive root to obtain
f 1 : x
1  1  x , x  , x  1.
If f is a one-one function, the graphs of
y  f ( x) and y  f 1 ( x) are reflections of
each other in the line y  x.
Do Misc. Exercise 11, pp. 172-173