Transcript Composite Functions

COMPOSITION
FUNCTIONS
“SUBSTITUTING ONE FUNCTION INTO ANOTHER”
The Composition
Function
 f  g x  f g x
Must be
able to
work either
This is read “f composition g” and means to copy the f
function down but where ever you see an x, substitute in
the g function.
f x   2 x  3
2

g x   4 x  1
3

2
f  g  2 4x 1  3
3
FOIL first and
then distribute
the 2
 32 x  16 x  2  3  32 x  16 x  5
6
3
6
3
g  f x  g  f x
This is read “g composition f” and means to copy the g
function down but where ever you see an x, substitute in
the f function.
f x   2 x  3
g x   4 x  1
2

3

3
g  f  4 2x  3 1
2
You could multiply
this out but since it’s
to the 3rd power we
won’t
 f  f x  f  f x
This is read “f composition f” and means to copy the f
function down but where ever you see an x, substitute in
the f function. (So sub the function into itself).
f x   2 x  3
g x   4 x  1
2

3

2
f  f  2 2x  3  3
2
A MathXTC Example of
Composite Functions
f x   x  5
g ( f ( x))  2 x  13x  14
2
g ( x)  ?
Try it !!
Method 1
f x   x  5
g f x   2 x  13x  14
2
g f ( x)  2 x  13x  14
2
g ( x  5)  2 x  13x  14
2
g ( x  5)  2( x  5)  ???
2
 2 x  20 x  50  ???
2
g ( x  5)  2( x  5)  7 x  36
2
 2( x  5)  7( x  5)  1
2
Method 1
f x   x  5
g f x   2 x  13x  14
2
g f ( x)  2 x  13x  14
2
g ( x  5)  2 x  13x  14
2
g ( x  5)  2( x  5)  ???
2
 2 x  20 x  50  ???
2
g ( x  5)  2( x  5)  7 x  36
2
 2( x  5)  7( x  5)  1
2
Method 1
f x   x  5
g f x   2 x  13x  14
2
g ( x  5)  2( x  5)  7 x  36
2
 2( x  5)  7( x  5)  1
2
 g ( x)  2 x  7 x  1
2
Method 2
f x   x  5
g f x   2 x  13x  14
2
Let y  f ( x)  x  5
 x  y5
g f ( x)  2 x  13x  14
2
g ( y)  2 x  13x  14
2
g ( y )  2( y  5)  13( y  5)  14
2
 2 y  20 y  50  13 y  65  14
2
 2 y  7 y 1
2
Method 2
f x   x  5
g f x   2 x  13x  14
2
g ( y )  2( y  5)  13( y  5)  14
2
 2 y  20 y  50  13 y  65  14
2
 2 y  7 y 1
2
 g ( x)  2 x  7 x  1
2
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au