Răspuns:
1.a
f(x)={sinx²/2x x<0
{sinx x≤0
Ls x->0 x<0 limsinx²/2x=1/2*lim sinx²/x=1/2limsinx²/x²*x=
1/2lim(sinx²/x²)*x=1/2*1*0=0
Ld x->0 limsinx=sin0=0
f(0)=sin0=0
Ls=Ld=f(0)=0 => f continua in 0
b)f(x)={x²+3 x<1
{2ˣ-2 1≤x<2
{x+2 x≥2
Continuitatea in 1
Ls x->1,x<1lim(x²+3)=1²+3=4
Ldx->1,x.>1lim(2ˣ-2)=2¹-2=0
4=/=o=> f nu este continua in 1
Continuitatea in 2
Ls x->2 x<2 lim (2ˣ-2)=2²-2=4-2=2
Ld x->2 x>2 lim(x+2)=2+2=4
4=/=2 => f nu este continua in 2
Ex2
f(x)={x²+a² x∈(-∞.a]
{3x-1 x∈(a,∞)
Ls x->a x<a lim(x²+a²)=a²+a²=2a²
Ld x->a x>a lim (3x-1)=3a-1
f(a)=2a²
Pui conditia 2a²=3a-1
2a²-3a+1=0
a1=1
a2= -1/2
a={-1/2<1}
Ex3
f:[1,7]→R
f(x)=x²-3x-4
f(1)=1-3-4=-6<0
f(7)=7²-3*7-4=49-21-4=24>0
Functia ia valori de semne contrare in capatul intervalului, deci se anuleaza in acest interval
x²-3x-4=0
x1=-1
x2=4
{-1,4}∈[-6,24]
b)f(x)=x²+2x-8
x²+2x-8=0
x1=-4, x2=2
{-4,2}}∈[-6,24]
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