Question

Let f(x) = ax + b where a and b are real numbers, f(f(f(1))) = 29 and f(f(f(0))) = 2. Then b equals ??
A start would be good.

Answer 1

help yourself xD

Answer 2

:/

Answer 3

Sarah, if you're not going to help, at least don't insult the person asking.

Answer 4

umm. i just realised.. shouldnt i just find f(f(f(x))) ??

Answer 5

Haha. Its ok Trash.

Answer 6

trashman its none of your business, he knows im joking so be quiet and dont you dare tell me what to do..

Answer 7

f(f(f(x)))=a^3(x) + a^2b + b .. is that right?

Answer 8

I have \(\large a^3+a^2b+ab+b \implies 29\)

Answer 9

Oh yeah. I'm sorry, i messed it up. ^^

Answer 10

Ok this should be easy now, ig? making factors and dividing the two equations? is that where we are going?

Answer 11

I'd think so, yes

Answer 12

Its not working out like that.. @jdoe0001

Answer 13

a sec lemme check about

Answer 14

That's a lot of f's o-O

Answer 15

Yeah its like. f(f(x))= a (ax+b) + b.. you just keep putting the value of f(x) in the x..

Answer 16

That sounds very confusing.. although f(x) is just a fancy way of saying y

Answer 17

@ohyeahh what did you get for f(f(f(0))) ?

Answer 18

a^2.b + ab +b = 2 .. thats right?

Answer 19

yes

Answer 20

now gimme a sec to paste, tis a system of equations, so you know :)

Answer 21

Take your time. Thank you.

Answer 22

f(x)=ax+b
f(1)=a+b
f(f(1))=f(a+b)=a(a+b)+b=a^2+ab+b
f(f(f(1)))=a(a^2+ab+b)+b=a^3+a^2b+ab+b=29
f(0)=0*a+b=b
f(f(0))=f(b)=ab+b
f(f(f(0)))=f(ab+b)=a(ab+b)+b=a^2b+ab+b=2
a^3+a^2b+ab+b=29
a^2b+ab+b=2
a^3+2=29
a^3=29-2=27
a=3
(3^2+3+1)b=2
b=2/13

Answer 23

$$
a^3+a^2b+ab+b=29\\
0+a^2b+ab+b=2 \ \ (\times \color{red}{-1})\\
--------------\\
a^3+a^2b+ab+b=29\\
0\color{red}{-a^2b-ab-b}=-2\\
---------------\\
a^3=27 \implies a=\sqrt[3]{27} = 3
$$

Answer 24

I guess finding a and then finding b should be easier?
I really couldnt follow how you found b. I got confused by the time i reached the end.

Answer 25

and from there, as you can see @surjithayer lines, you substitute (a) to get (b)

Answer 26

lol sorry.. it was you who wrote that. it was surjith.. but your method looks easier. i mean i couldnt really follow him..

Answer 27

not you* ughh .