# composite functions 2

• Oct 10th 2009, 12:29 PM
Oasis1993
composite functions 2
Hello everyone,
Here are some questions.I dont have the answers to them and im kind of having trouble. I would really appreciate it if you showed me the way of solving A and just the answers to the rest because i want to solve them myself.

Given that f(x)= 5-x and g(x)= 4/x , where x is all real numbers except for 0 and 5 find the value

a) gg(1)
b) fffff(6)
c) fggf(2)

Thank you
• Oct 10th 2009, 12:48 PM
skeeter
Quote:

Originally Posted by Oasis1993
Hello everyone,
Here are some questions.I dont have the answers to them and im kind of having trouble. I would really appreciate it if you showed me the way of solving A and just the answers to the rest because i want to solve them myself.

Given that f(x)= 5-x and g(x)= 4/x , where x is all real numbers except for 0 and 5 find the value

a) gg(1)
b) fffff(6)
c) fggf(2)

Thank you

g(x) = 4/x

g(1) = 4/1 = 4

g(4) = 4/4 = 1

so ...

g[g(1)] = g[4] = 1
• Oct 10th 2009, 01:08 PM
Oasis1993
thank you.
would i have to insert 6, 5 times?
I dont really get how to?
• Oct 10th 2009, 01:18 PM
skeeter
Quote:

Originally Posted by Oasis1993
thank you.
would i have to insert 6, 5 times? no, that's not how it works.

f(x) = 5-x

note that f(6) = 5 - 6 = -1 and f(-1) = 5 - (-1) = 6

work inside out ...

$\displaystyle f[f[f[f[f(6)] =$

$\displaystyle f[f[f[f(-1)] =$

$\displaystyle f[f[f(6)] =$

$\displaystyle f[f(-1)] =$

$\displaystyle f(6) = -1$

get the idea?
• Oct 10th 2009, 01:19 PM
e^(i*pi)
Quote:

Originally Posted by Oasis1993
Hello everyone,
Here are some questions.I dont have the answers to them and im kind of having trouble. I would really appreciate it if you showed me the way of solving A and just the answers to the rest because i want to solve them myself.

Given that f(x)= 5-x and g(x)= 4/x , where x is all real numbers except for 0 and 5 find the value

a) gg(1)
b) fffff(6)
c) fggf(2)

Thank you

f(x) = 5-x
ff(x) = 5-(5-x) = x
fff(x) = 5-x

Note how this repeats itself. If there is an odd number of f then it will be 5-x. Even numbers will be x.

fffff(x) = 5-x
• Oct 10th 2009, 01:31 PM
Oasis1993
Thank you very much!

so for question C

fggf(2)

would it be like this?

f(2)= 5-2=3
g(3)=4/3
g(4/3)=4/(4/3)
f(3.08)=5-3.08= 1.92

?
• Oct 10th 2009, 01:44 PM
e^(i*pi)
Quote:

Originally Posted by Oasis1993
Thank you very much!

so for question C

fggf(2)

would it be like this?

f(2)= 5-2=3
g(3)=4/3
g(4/3)=4/(4/3) e^(i*pi) = $\displaystyle \frac{4}{\frac{4}{3}} = 4 \times \frac{3}{4} = 3$

f(3.08)=5-3.08= 1.92

?

$\displaystyle g^{-1}(x) = \frac{4}{x} = g(x)$

Therefore wherever you get two lots of g next to each other they will cancel to leave x.

$\displaystyle fggf(x) = ff(x) = x$ as established in question 2.

Therefore $\displaystyle fggf(2) = 2$

The function $\displaystyle h(x) = \frac{a}{x} \: , \: a,x \neq 0$ is equal to it's inverse
• Oct 10th 2009, 01:47 PM
skeeter
Quote:

Originally Posted by Oasis1993
Thank you very much!

so for question C

fggf(2)

would it be like this?

f(2)= 5-2=3
g(3)=4/3
g(4/3)=4/(4/3)
f(3.08)=5-3.08= 1.92 ... no

?

put away the calculator ...

4/(4/3) = 4(3/4) = 3

f(3) = 2
• Oct 10th 2009, 01:50 PM
Oasis1993
Thank you!!