Exercise for young people (2), Analysis

[Steven11]

Not too difficult, actually.

Consider the following function
$f(x)=x^2+ax+b$

i) Prove that there is $x_0 in[-1,1]$ such that $|f(x_0)|>=1/2$

ii) Prove that if $|f(x)|<=1/2 \quad \quad forallx in[-1,1]$
it must be: $f(x)=x^2-1/2$

Have a good work.

[Fioravante Patrone1]

"fu^2":darboux th: let $f:I\subRR->RR$ a function continuos in I and $f(x)!=c$ (non constant value ). Then f(I) is an interval and if $I=[a,b]$, $f(I)=[m,M]$ with $m=minf$ and $M=maxf$.

Theorem (Darboux?). Let $f:I\subRR->RR$ be continuous. Then, f(I) is an interval.

Remark. Singletons and the empty set are intervals, so the additional condition that $f$ is not constant is pointless.

Corollary. Under the assumptions of the theorem above, if $I=[a,b]$, then $f(I)=[m,M]$, where $m=minf$ and $M=maxf$.
Prrof. The proof of the corollary is immediate, thanks to the Weierstrass' theorem.

Remark. Usually, the "Darboux's theorem" refers to the interesting fact that the derivative of a function has the intermediate value property, even if it is not continuous. The theorem that has been named here as "Darboux's theorem" is usually known as the "Intermediate value theorem."

[fu^2]

"Steven":[quote="fu^2"]
in this case if the length of the interval is 1, then the absolute value of minf or maxf is >=1/2. So the number of $x_0$ is least 2! why?... bye!

I do not understand this sentence, actually.
Why do you say "why" ? [/quote]

"why" because the proof of this fact is an exercise for other people that if you read my proof, this sentence is a corollary (if corollario in english is corollary ) (if this sentence is correct... )

but do you understand the sentence?

[Steven11]

Of course, thanks for your post.
Corrections are precious, always.

Have a good Sunday!

[bryce1]

I just read the title "Exercise for young peaple (2), Analysis" ... and I suggest a first correction firtsly: what is "peaple"???? Should it be "people"?

Regarding the exercise... I'll read better whole the thread and I'll tell you. This post was for letting know you about the usage of word "peaple". I think you meant the "group of persons" and not the genealogy meaning...

Cheers!

[Steven11]

"fu^2":
in this case if the length of the interval is 1, then the absolute value of minf or maxf is >=1/2. So the number of $x_0$ is least 2! why?... bye!

I do not understand this sentence, actually.
Why do you say "why" ?

[fu^2]

darboux th: let $f:I\subRR->RR$ a function continuos in I and $f(x)!=c$ (non constant value ). Then f(I) is an interval and if $I=[a,b]$, $f(I)=[m,M]$ with $m=minf$ and $M=maxf$.

In other word the Darboux th say that f assume all value between m and M, in this case if the length of the interval is 1, then the absolute value of minf or maxf is >=1/2. So the number of $x_0$ is least 2! why?... bye!

[Steven11]

it's ok?

I think so.
I say "I think" because I actually do not know the "Darboux Th.", but the second part seems good to me.

As a matter of fact, I solved the problem in another way, which partially looks like the yours, but uses only some elementary considerations of basic Algebra.
It could be thought by a student who studied Algebra of liceo and the parabola.

I will write it, if nobody does it. I wait for some days.

Bye fu^2

[fu^2]

the function $f(x)$ is continuos in [-1,1].

so we'll obtain the result if the length $|maxf(x)-minf(x)|>=1$ with $f:[-1,1]->RR$ (the max and the min exists for the weiestrass' theorem), in fact for the darboux's theorem $EEx_0$ tc $|f(x_0)|>=1/2$.

the small value of f is in $barx=-(a/2)$. So $x=-a/2$ is the axis of f, we can change the coordinate system: $x'=x+a/2$ so the equation became: $f(x)=(x-a/2)^2+a(x-a/2)+b=x^2+a^2/2+b=>f(-1)=f(1)=1+a^2/2+b$ and $f(0)=a^2/2+b$ is the small value of f. so $f(1)=maxf(x)$ and $f(0)=minf(x)$ => $f(1)-f(0)=1+a^2/2+b-(a^2/2+b)=1$.


if $|f(x)|<=1/2=>$ the $maxf=1/2$ and $minf=-1/2$("in quanto"(i don't know how can say on English) the length is always 1!), using the new parabola we obtain that $1+a^2+b=1/2$ and $a^2/2+b=-1/2$ and the result $a=0$and $b=-1/2$

it's ok?