Question

Se considera functia f:R – R, f(x) = x^2-x+1/x.
(continuarea se afla in poza)

Va rog muuult!​

Answer 1

Răspuns:

Explicație pas cu pas:

a) x -f(x) = x - (x^2 -x+1)/x = (x^2 -x^2 +x -1)/x =

(x-1)/x = (1 -1/x)

L = lim x->inf(1-1/x)^x,  logaritmam:

lnL =  lim x->inf( ln(1-1/x)^x) =

lim x->inf(lnx*(1-1/x))  =

lim x->inf(ln(1-1/x))/(1/x) = 0/0, aplicam l'Hopital:

lim x->inf(ln(1-1/x))/(1/x) =lim x->inf (1/x^2*(-x^2))/(1 -1/x)=

lim x->inf( -1/(1-1/x) = -1/(1-0) = -1

Deci lnL = -1 = ln(1/e),  L = 1/e

b) f'(x) = ((2x-1)x -(x^2 -x +1)*1)/x^2 =

(2x^2 -x -x^2 +x -1)/x^2 = (x^2 -1)/x^2

f'(x) = 0, x^2-1=0,  x1 = -1,  x2 = 1

lim x->-inf f(x) = -inf, lim x->+inf f(x) = +inf

f(-1) = (1+1+1)/-1 = -3,  f(1) = 3/1 = 3

In x= 0, avem nedeterminare, deci asimpt. vert.

Limita la st. in x=0 din f(x) = 1/-0 = -inf

Limita la dr. in x=0 din f(x) = 1/+0 = +inf

(-1; -3) punct de max,  (1; 3) punct de minim

x in (-inf, -1) crescatoare

x in (-1, 0)  descrescatoare

x in (0, 1) descrescatoare

x in (1, +inf) crescatoare

c) y = mx +n

m = lim x->inf(f(x)/x) = lim x->inf(x^2 -x +1)/x^2 = 1

n = lim x->inf(f(x) -x) =lim x->inf(x^2 -x +1)/x -x)=

lim x->inf(x^2 -x +1 -x^2)/x = lim x->inf(-x +1)/x = -1

y = x -1  asimpt. oblica la +inf