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

-------------------------

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

_____________________________________________________-

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

------------------------------------------------

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: