Question

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Answer 1

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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