Cambridge Additional Mathematics

50 Functions (Chapter 2)

Algebraically, if f(x)=x^4 and g(x)=2x+3 then

(f±g)(x)=f(g(x))

=f(2x+3) fgoperates onxfirstg

=(2x+3)^4 ffoperates ong(x)nextg

and (g±f)(x)=g(f(x))

=g(x^4 ) ffoperates onxfirstg

=2(x^4 )+3 fgoperates onf(x)nextg

=2x^4 +3

So, f(g(x)) 6 =g(f(x)).

In general, (f±g)(x) 6 =(g±f)(x).

We can also compose a functionfwith itself. The resulting function is (f±f)(x) or f^2 (x).

In general, (f±f)(x) 6 =(f(x))^2.

Example 12 Self Tutor

Given f:x 7! 2 x+1 and g:x 7! 3 ¡ 4 x, find in simplest form:

a (f±g)(x) b gf(x) c f^2 (x)

f(x)=2x+1 and g(x)=3¡ 4 x

a (f±g)(x)

=f(g(x))

=f(3¡ 4 x)

= 2(3¡ 4 x)+1

=6¡ 8 x+1

=7¡ 8 x

b gf(x)

=g(f(x))

=g(2x+1)

=3¡4(2x+1)

=3¡ 8 x¡ 4

=¡ 8 x¡ 1

c f^2 (x)

=f(f(x))

=f(2x+1)

= 2(2x+1)+1

=4x+2+1

=4x+3

In the previousExampleyou should have observed how we can substitute an expression into a function.

If f(x)=2x+1then f(¢) = 2(¢) + 1

) f(3¡ 4 x) = 2(3¡ 4 x)+1.

Example 13 Self Tutor

Given f(x)=6x¡ 5 and g(x)=x^2 +x, determine:

a (g±f)(¡1) b (f±f)(0)

a (g±f)(¡1) =g(f(¡1))

Now f(¡1) = 6(¡1)¡ 5

=¡ 11

) (g±f)(¡1) =g(¡11)

=(¡11)^2 +(¡11)

= 110

b (f±f)(0) =f(f(0))

Now f(0) = 6(0)¡ 5

=¡ 5

) (f±f)(0) =f(¡5)

=6(¡5)¡ 5

=¡ 35

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Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_02\050CamAdd_02.cdr Thursday, 19 December 2013 2:38:52 PM BRIAN