Răspuns:
derivabilitatea in xo=0
x->0 lim f(x)-f(xo)/(x-x0)=
lim(2x-x²-2*0+0²)/(x-0)=
lim (2x-x²)x=lim(2-x)=2-0=2
Limita in 0 exista si este finta.Functia este derivabila in 0
xo=1
x-1 lim[2x-x²)-(2*1-1²)]/(x-1)=
lim(2x-x²-2+1)/(x-1)=
lim(-x²+2x-1)/(x-1)=lim-(x²-2x+1)/(x-1) =
-lim(x-1)²/(x-1)= -lim(x-1)=
-(-1-1)=2
Limita finita.Functia derivabila in 1
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x=2
x->2 lim[(2x-x²)-(2*2-2²)]/(x-2)=
lim(2x-x²-4+4)/(x-2)=
lim(-x²+2x)/(x-2)=
-limx(x-2)/(x-2)=
-limx= -2
Limita exista , f derivabila
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Ecuatia tangentei
y-yo=f `(xo)(x-xo)
xo=0
yo=f(0)=2*0-0²=0
yo=0
f `(x)=2-2x
f `(0)=2-2*0=2
Faci inlocuirile
y-0=2(x-0)
y=2x
tangenta in xo=1
f(1)=2*1-1²=2-1=1=yo
f `(1)=2-2*1=2-2=0
deci ecuatia
y-1=0*(x-1)
y-1=0
y=1
-----------------
xo=2
yo=f(2)=2*2-2²=4-4=0
f `(2)=(2-2*2)= -2
ecuatia devine
y-0=2(x-2)
y=2x-4
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f(x)=x³
derivabilitate in xo=1
x->1 lim[f(x)-f(1)]/(x-1)=
lim(x³-1³)/(x-1)=
lim(x-1)(x²+x+1)/(x-1)=simplifici prin (x-1)
lim(x²+x+1)=1²+1+1=3
functia derivabila in xo=1
Ecuatia tangentei
y-yo=f `(xo)(x-xo)
yo=f(1)=1³=1
f `(x)=(x³) `=3x²
f `(1)=3·1²=3
inlocuiesti
y-1=3(x-1)
y=3x-3+1
y=3x-2
Explicație pas cu pas:
