Something serius

[Anonyme]

Sont donnees les fonctions derivable,avec df(x)/dx continue et la formule: f(x)=x^2-Integ(2x-c)dx-2005, c est un Reel
i)Demontrer que ces fonctions n'ont ni maximum ni minimum.
ii)Demontrer qu'il existe f^-1(x)

[leibniz]

Salut,
qu'est ce que tu veux dire par integr(2x-c)? Est ce que c'est la primitive de 2x-c?
Merci

[Anonyme]

i mean Integral.I dont know very good french.Exemple:Integ x dx=(x^2)/2+ C, C est un reel.

[Anonyme]

generaly Integ f '(x)dx=f(x) + C, C est un reel

[leibniz]

Hello,
Je crois que dans ce cas la fonction f sera definie par : f(x)=-x²+2cx-2005+c'
Il sagit dans ce cas d'une parabole qui admet un maximum puisque le coefficient de x² est negatif.
Essayer d'etre sérieux un peu - Try to be serious.
Merci

[Anonyme]

It appears that

f(x)=cx+(d-2500), where c and d are real numbers,

which is the equation of a straight line. The function f has therefore no local minima and no local maxima (unless I really got it wrong?). The max and min of f on a finite interval [t0, t1] depends on the sign of c: if c>0, the min is at t0 and the max at t1, if c<0 it's the other way around.

As for the inverse function, well, I'd say f^-1(x)=(y-d+2005)/c.

SYL.

[Anonyme]

that's right.Gongratulations!!!!!!!!

[Anonyme]

I have tell you that the fonction f(x) of this problem was for all x in R so the straight line f(x)=cx+(d-2005) have not min or max.(f(D)=(-inf,+inf)
with inf i mean infinie.

[Anonyme]

that was to be shown, wasn't it ?

f^-1(x) exists however, unless c=0.

When c=0, f is a constant (f(x) = d-2005) so f is not, well, invertible ? (I don't know english very well, sorry).

But if c isn't 0, f^-1(x) = (2005-d -x)/c

[Anonyme]

soit f(x)=x^2-integ(2x-c)-2005
Donc df(x)/dx=2x-2x+c=c ,c is one real number
But f(x)=x^2-integ(2x-c)-2005 is for all x ion R and df(x)/dx=c is olso for all x on R.
c can be only positive or negative but no 0.
Because if c=0 it must be df(x)/dx=o for all x on R.
But if df(x)/dx=o for all x on R, f(x) must be constant, but i tell you that f(x) isn't constant,and that mean that df(x)/dx mustn't be 0.(this isn't difficult to show).
So df(x)/dx>0 if c>0 for all x on R and like this f(x) est strictement croissante on R (1)
If df(x)/dx<0 if c<0 for all x on R and like this f(x) est strictement decroissante on R.(2)
So there isnt max or min.
(It isn't difficult to show (1) and (2))
If you want i can show you.Ask me.

[Anonyme]

Yes my nigger,me to i dont know veru well english.I am from Greece.