f(x) = x3 +x²-x+3
d/dx f(x) = 3x²+2x-1
3x²+2x-1 = 0
del. = 4+12=16
p.del = 4
x1= -2+4 / 6 = 1/3
x2= -2-4 /6 = -1
rosnie (-∞;-1) u (1/3 +∞)
maleje (-1 ; 1/3)
Ekstrema f(x) (x²+3x)/(x-1) Df=R/{1}
d/dx f(x) = (2x+3)(x-1) - (x²+3x)(1) / (x-1)²
L=2x²+x-3 -x²-3x = x²-2x-3
M - zawsze >= 0 wiec nie biore go pod uwage
x²-2x-3 = 0
del=4+12=16
p.del=4
x1=3
x2= -1
2pkty podejrzane o ekstrema, sprawdzamy z Df
oba pasuja, ekstrema f(x) to x1=(3,9) x2=(-1;1)