Răspuns:
f(x)=x-ln(x²+1)
a) f `(x)= 4x `-2x/(x²+1)=
4-2x/(x²+1)
b)f(x+1)=4(x+1)-ln[(x+1)²+1]
lim [4(x+1)-ln((x+1)²+1)-4x+ln(x²+1)=4x+4-4x-ln((x+1)²+1)+ln(x²+1)=
4 +ln(x²+1)-ln((x+1)²+1)=
4+ln(x²+1)/(x²+2x+1+1)=4+ln(x²+1)/(x²+2x+2)
x->∞lim[4+ln(x²+1)/(x²+2x+2)]=
4+limLn(x²+1)/(x²+2x+2)=
4+ln lim(x²+1)/(x²+2x+2)=
4+ln1=4+0=4
c)nu se poate demonstra atat timp cat nu se cunoaste domeniul si codomeniul functiei
Explicație pas cu pas:
