Question

Help please!!!(:
A number a is called a fixed point of a function f if f(a) = a. Prove that if f'(x) does not equal 1 for all real numbers x, then f has at the most one fixed point.

Hint: Consider the function g(x)=f(x)-x.

Answer 1

Since part of the hypothesis is that f'(x) is never equal to 1, that sorta hints that maybe you want to consider g'(x). If g(x) = f(x)-x, what is g'(x)?

Answer 2

ummmm? haha

Answer 3

g'(x) can't equal 1 either? @joemath314159

Answer 4

\[(g(x))'=(f(x)-x)'=f'(x)-1\]

Answer 5

If f'(x) can't equal 1, then g'(x) cant equal.....

Answer 6

f(x)?

Answer 7

>.< Trying to type enough as to not total give the answer away is hard lol.

Answer 8

Lets see. Let me put it this way. We have:\[g'(x)=f'(x)-1.\]If f'(x)=1, then what would g'(x) be?

Answer 9

f'(x)-1

Answer 10

I'm sorry but I'm so confused on this kind of stuff.

Answer 11

Thats right, but now f'(x) = 1. So it would be 1 - 1.

Answer 12

so g'(x) =0

Answer 13

Right! So if f'(x) = 1, then g'(x) = 0. But by our hypothesis, f'(x) is never equal to 1! So that means g'(x) is never equal to.....

Answer 14

g'(x) cant = 0

Answer 15

right right! alright, thats the first thing to notice. That g'(x) can never be zero.

Answer 16

Second thing to notice:

Lets say that "a" is a fixed point of f. That is:\[f(a)=a.\]Then what is g(a)?\[g(a)=f(a)-a=?\]

Answer 17

g(a)=f?

Answer 18

\[g(a)=f(a)-a\]but since f(a)=a, we can substitute:\[f(a)-a=a-a=0.\]So:\[g(a)=0.\]

Answer 19

ohhh, haha, I blanked for a bit.

Answer 20

So thats the second thing to notice, that if "a" is a fixed point of f (that is, f(a)=a), then g(a)=0.

Answer 21

Now lets keep our eyes on the prize. We want to prove that there is only one fixed point for f. If we have only one fixed point for f, that means there is only one solution for g(x)=0. Does that make sense?

Answer 22

yes, makes sense (:

Answer 23

Ok. now heres the last piece of the puzzle. Since g'(x) can never be zero (the first thing we noticed), that means g'(x) is always positive, or always negative. Let's say g'(x) is always positive. What can you conclude about g(x)? Is it possible for g(x) to increase some, then decrease some, if g'(x) is always positive?

Answer 24

Gotta think back to that Calculus lol.

Answer 25

if it's positive, it'll always be positive.

Answer 26

Right, if g'(x) is positive, then g(x) will always be increasing positively, never decreasing. Its always gotta be moving up. Now picture g(x) on the x-y plane. If its always moving up, is it possible for g(x) to cross the x-axis more than once?

Answer 27

no, since it's always increasing it'll once hit the x-axis once and never come back down

Answer 28

Correct! So if g(x) can only cross the x-axis once, how many solutions to the equation g(x)=0 can there be?

Answer 29

only one?

Answer 30

Thats right! Now look back to when I said "Now lets keep our eyes on the prize".

Answer 31

so because f'(x) cant equal 1 and g'(x) cant equal zero. with proof using the the fuction. we know that if f(x) is a neg or positive it will always either alwys be increasing or decreasing hitting the x-axis once, have only one solution(fixed point)

Answer 32

I know you have the right idea, but the way you're saying it is a little off.

Since f'(x) can't equal 1, it follows that g'(x) can't equal 0. This implies that g'(x) is either always positive or always negative. Thus g(x) will always be either increasing or decreasing, and can hit the x-axis at most one time. Therefore the equation g(x)=0 has at most one solution, and we conclude that f(x) has at most one fixed point.

Answer 33

tht's kinda sorta what I ment haha

Answer 34

lolol. Well, im off to bed, its almost 3 am here. Good luck with the rest of your work :)

Answer 35

Thank you for all the help! I really appreciated it! Night! (: @joemath314159