Finding fix points with two curves given

Asked Jan 25 '20 at 18:27

Active Jan 27 '20 at 18:46

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I am having trouble solving for $x$ (ie. find the fixed points) when two curves are given. I need to show that if $f: \mathbb R \to \mathbb R$ is a cubic function $f(x) = ax^3+bx^2+cx+d, ~ a \ne 0$ then $f$ always has a fixed point (hint use $g(x) = f(x) - x$)

I got up to

$ax^3+bx^2+cx+d = x$ ; set them to equal

$ax^3+bx^2+cx+d - x = 0$

$x (ax^2+bx+c-1) - d = 0$

edited Jan 25 '20 at 18:29

Simply Beautiful Art

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asked Jan 25 '20 at 18:27

computer891

1

$g$ is a polynomial of odd degree, so it has a real root... – Simply Beautiful Art Jan 25 '20 at 18:31
can you further elaborate? – computer891 Jan 25 '20 at 19:47
Proof that every polynomial of odd degree has one real root. – Simply Beautiful Art Jan 25 '20 at 19:51
but which comment follows the problem I'm doing? I'm confused what is being explained in this link. – computer891 Jan 25 '20 at 20:06
As the linked question shows, every odd degree polynomial has a real root. That is to say, there exists $x$ where $g(x)=0$. If you're still confused, you need to be more specific as to what you are confused with. – Simply Beautiful Art Jan 25 '20 at 20:09
I'm confused on why we use g(x)=f(x)−x as a clue. I'm assuming I need to set both the linear function and cubic function to equal each other and work out x, but i'm having trouble solving it further. Unless the approach i am doing is incorrect and there's another way – computer891 Jan 25 '20 at 20:11
What happens to $g(x)$ if $x=f(x)$? – Simply Beautiful Art Jan 25 '20 at 20:19
I'm not sure. Graphically I guess there would be some fixed point – computer891 Jan 25 '20 at 20:20
If $x=f(x)$, surely there is something you can conclude about $f(x)-x$. – Simply Beautiful Art Jan 25 '20 at 20:27

Finding fix points with two curves given

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