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Polynomial Function f

evinda

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MHB Site Helper
Apr 13, 2013
3,720
Hi!!! I have also an other question (Blush)

Knowing that $f$ is a polynomial function,how can I show that there is a $y \in \mathbb{R}$,such that $|f(y)|\leq |f(x)| \forall x \in \mathbb{R}$ ?
 
Last edited:

ZaidAlyafey

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MHB Math Helper
Jan 17, 2013
1,667
Are you sure of the question? how can all polynomials be bounded below by a positive real number ?
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Are you sure of the question? how can all polynomials be bounded below by a positive real number ?
I forgot the absolute value at $f(x)$.Sorry!! :eek:
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Let us take two cases

[1] If $f$ intersects the x-axis at \(\displaystyle y=a\) then since $|f(a)|=0$ we have $|f(x)| \geq |f(a)|$.

[2] If $f$ doesn't intersect the x-axis then it is a polynomial of even degree and you can find the minimum of \(\displaystyle |f|\) by differentiation.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Let us take two cases

[1] If $f$ intersects the x-axis at \(\displaystyle y=a\) then since $|f(a)|=0$ we have $|f(x)| \geq |f(a)|$.

[2] If $f$ doesn't intersect the x-axis then it is a polynomial of even degree and you can find the minimum of \(\displaystyle |f|\) by differentiation.
How can I use these facts to show that there is such a $y$ ?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
What Zaid is referring to is that any real polynomial of odd degree has a root.

If our polynomial has a root, say at $a$, then $f(a) = 0$, so we may take $y = a$.

This takes care of all polynomials that have a real root (which includes ALL polynomials of odd degree, and SOME polynomials of even degree).

Which leaves with with even degree polynomials that have no real root (like, for example, $x^4 + 1$).

If $f$ is of even degree, then $f'$ is of odd degree. By the discussion above, $f'$ has a root (it may have more than one).

Now $f$ is bounded below by 0 (if the leading term's coefficient is > 0) or bounded above by 0 (if the leading term's coefficient < 0). Since we are considering $|f|$ it doesn't matter if we talk about $f$ of $-f$, since both have the same absolute value.

So we may as well assume $f > 0$. We know that the set of real roots of $f'$ is non-empty. We can (if we feel like being thorough) distinguish 3 cases:

1. $f'$ has just one real root. This must be a global minimum for $f$.

2. $f'$ has two real roots. One of these must be the global minimum, and the other an inflection point.

3. $f'$ has 3 real roots. Two of these are local minima, the third (which is between the other two) is a local maximum. The local minimum with the smallest value is the desired global minimum.

In all 3 cases, a global minimum exists, which we can then choose to be our $y$.

As for our example above, we find that $f'(x) = 4x^3$, which has the sole root $x = 0$.

And, from inspection, it is not hard to see to $x^4 + 1$ has a minimum value of 1 at $x = 0$.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
What Zaid is referring to is that any real polynomial of odd degree has a root.

If our polynomial has a root, say at $a$, then $f(a) = 0$, so we may take $y = a$.

This takes care of all polynomials that have a real root (which includes ALL polynomials of odd degree, and SOME polynomials of even degree).

Which leaves with with even degree polynomials that have no real root (like, for example, $x^4 + 1$).

If $f$ is of even degree, then $f'$ is of odd degree. By the discussion above, $f'$ has a root (it may have more than one).

Now $f$ is bounded below by 0 (if the leading term's coefficient is > 0) or bounded above by 0 (if the leading term's coefficient < 0). Since we are considering $|f|$ it doesn't matter if we talk about $f$ of $-f$, since both have the same absolute value.

So we may as well assume $f > 0$. We know that the set of real roots of $f'$ is non-empty. We can (if we feel like being thorough) distinguish 3 cases:

1. $f'$ has just one real root. This must be a global minimum for $f$.

2. $f'$ has two real roots. One of these must be the global minimum, and the other an inflection point.

3. $f'$ has 3 real roots. Two of these are local minima, the third (which is between the other two) is a local maximum. The local minimum with the smallest value is the desired global minimum.

In all 3 cases, a global minimum exists, which we can then choose to be our $y$.

As for our example above, we find that $f'(x) = 4x^3$, which has the sole root $x = 0$.

And, from inspection, it is not hard to see to $x^4 + 1$ has a minimum value of 1 at $x = 0$.
I understand..Thank you!! :)
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Alternatively, you can prove that for every polynomial $f(x)$ there exist points $M,N$ such that $|f(x)|>f(M)$ for all $x<M$ and $|f(x)|>f(N)$ for all $x>N$. The boundedness theorem says that $|f(x)|$ has a minimum in $[M,N]$, so the global minimum is the least of that minimum, $f(M)$ and $f(N)$. This approach does not require using the derivative.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Alternatively, you can prove that for every polynomial $f(x)$ there exist points $M,N$ such that $|f(x)|>f(M)$ for all $x<M$ and $|f(x)|>f(N)$ for all $x>N$. The boundedness theorem says that $|f(x)|$ has a minimum in $[M,N]$, so the global minimum is the least of that minimum, $f(M)$ and $f(N)$. This approach does not require using the derivative.
I understand..Thanks a lot!!! :)