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(a) ( f o h ) ( x )
Not much to do here other than run through the formula.
© 2007 Paul Dawkins 193 http://tutorial.math.lamar.edu/terms.aspx College Algebra ( f o h) ( x) = f éh ( x )ù
= f é x + 1ù
= ( ) 2 x +1 - 3 = x +1- 3
[Return to Problems] (b) ( h o f ) ( x) Again, not much to do here. ( h o f ) ( x ) = h é f ( x )ù
= h é x 2 - 3ù
= x2 - 3 + 1
= x2 - 2
[Return to Problems] (c) ( f o 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 o f ) ( x) = f é f ( x )ù
= f é x - 3ù
û = ( x 2 - 3) - 3
2 = x4 - 6 x2 + 9 - 3
= x4 - 6 x2 + 6 [Return to Problems] (d) ( h o 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 o h ) ( 8 ) = h é h ( 8) ù
û = h é 8 + 1ù
= hé 9ù
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.
- Spring '12