1. ## 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

2. 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

3. thank you.
would i have to insert 6, 5 times?
I dont really get how to?

4. 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 ...

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

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

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

$f[f(-1)] =$

$f(6) = -1$

get the idea?

5. 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

6. 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

?

7. 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) = $\frac{4}{\frac{4}{3}} = 4 \times \frac{3}{4} = 3$

f(3.08)=5-3.08= 1.92

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

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

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

Therefore $fggf(2) = 2$

The function $h(x) = \frac{a}{x} \: , \: a,x \neq 0$ is equal to it's inverse

8. 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

9. Thank you!!