Question

a,b=? astfel incat f(x)= x²+ax+b/x²+1,are punct de extrem=2 in punct de abscisa 1​

Answer 1

Răspuns:

f90x) ;(0,∞)

f'(x) =[2lnx * (1/x)*x-ln²x *1]/x²= (2lnx-ln²x)/x²= lnx( 2-lnx)/x²

derivata se anuleaza pt lnx=0; x=1; f(1)=0

si pt lnx=2 x=e²

f(x)=4/e²

f(1)=0, mimim local

f(e²)=4/e², maxim local

Explicație pas cu pas: