functie omografica
monotona
avem conditiile f(3)=-2
si f '(3)=1
pt ca 45de grade cu axa o x face prima bisectoare y=x care are panta 1
f'(x) = (m(x+3) -mx-n) /(x+3)²
f '(x)=(3m-n)/(x+3)²
deci
cele 2 ecuatii in m si n sunt
f(3)=(3m+n)/6=-2
si
f'(3)=(3m-n)/6²=1
sau 3m+n=-12
3m-n=36
adunam
6m=24
m=4
n=12-3m=-12-12=-24
m=4
n=-24
f(x)=(4x-24)/(x+3)
verificare
f(3)=-12/6=-2 adevarat
f'(x)=(4(x+3)-4x)/(x+3)²=12/(x+3)²
f'(3)=(12-(-24))/36=36/36=1, adevarat,
adevarate ambele, bine rezolvat
