# Thread: Find equation w/ 3 Fixed points

1. ## Find equation w/ 3 Fixed points

Hello

I am returning to school after a long break. One of my assignments is a mix of ODE and discrete math and is clogging my mind.

I have been struggling on my 12 question homework for 4 hours now and have made it 1/4 of the way through. Luckily i have found this forum and hope someone(s) can help.

The question is this:
Find a continuous function:
f : [0, 1] --> [0, 1] with exactly three fixed points and draw its graph.

The fixed points is throwing me for a loop.

Any suggestions??

2. Originally Posted by npthardcorebmore
Hello

I am returning to school after a long break. One of my assignments is a mix of ODE and discrete math and is clogging my mind.

I have been struggling on my 12 question homework for 4 hours now and have made it 1/4 of the way through. Luckily i have found this forum and hope someone(s) can help.

The question is this:
Find a continuous function:
f : [0, 1] --> [0, 1] with exactly three fixed points and draw its graph.

The fixed points is throwing me for a loop.

Any suggestions??
Start with definition: x is a fixed point of a function f if and only if f(x) = x.

All you have to do is draw a continuous function in the region $x \in [0,1], y \in [0,1]$ that intersects with the line y=x three times. If you start at the point (0,0) and end at the point (1,1) then you can just have it cross the line y=x once for $x \in (0,1)$.

You can make it piecewise...

3. How about $f(x)=x^3$? We have $f(-1)=-1$, $f(0)=0$, $f(1)=1$. These are all of the fixed points since $x^3=x$ has only three solutions, by the Fundamental Theorem of Algebra.

4. Originally Posted by roninpro
How about $f(x)=x^3$? We have $f(-1)=-1$, $f(0)=0$, $f(1)=1$. These are all of the fixed points since $x^3=x$ has only three solutions, by the Fundamental Theorem of Algebra.
But when restricted to the domain [0,1], f(x)=x^3 only has two fixed points.

5. Originally Posted by undefined
But when restricted to the domain [0,1], f(x)=x^3 only has two fixed points.
Scale, $f\left(\frac{x+1}{2}\right)$

6. Originally Posted by Drexel28
Scale, $f\left(\frac{x+1}{2}\right)$
Maybe I'm misinterpreting.. $f(x) = x^3$ so $f\left(\frac{x+1}{2}\right)=\left(\frac{x+1}{2}\ri ght)^3$ which has only two fixed points for $x \in [0,1]$ given by $x \in \{-2+\sqrt{5},1\}$.

7. Originally Posted by undefined
Maybe I'm misinterpreting.. $f(x) = x^3$ so $f\left(\frac{x+1}{2}\right)=\left(\frac{x+1}{2}\ri ght)^3$ which has only two fixed points for $x \in [0,1]$ given by $x \in \{-2+\sqrt{5},1\}$.

Oh! fixed points! ....why don't you just construct a polygonal line that intersects the line $y=x$ three times?

8. Originally Posted by Drexel28
Oh! fixed points! ....why don't you just construct a polygonal line that intersects the line $y=x$ three times?
that would be a piecewise function correct?

i was just hoping for something a little more "pretty" than that

9. Originally Posted by npthardcorebmore
that would be a piecewise function correct?

i was just hoping for something a little more "pretty" than that
My original thought, before the piecewise idea, was to transform the sine curve so that it's rotated 45 degrees and can be stretched or compressed (along the line y=x) to produce as many fixed points as desired (greater than or equal to 2)... but since I figured you might ask how this is done, and I'd have to think about that, I just went the easier route.

10. would a piecewise function preserve continuity though?

11. Yes. It would preserve continuity but not differentiability in the "transition" points (you can see this by comparing the one-sided limits..)