Question

1) f(x)=cos +x^2/2. Atatati ca fx >=1.
2) f:(0,infinit )->(4,infinit ), fx =2^x+3, gx - inversa functiei f. g(5)=?

Answer 1

1)f'(x)=-sinx+x=0=>sinx=x=>x=0
x _______________0________________
f'(x)-------------------------1++++++++++++++
f(x) descrescatoare      crescatoare
x=0 este punct de minim adica f(x)≥f(0)=>f(x)≥1

2)Pentru a determina inversa functiei f, rezolvam ecuatia f(x)=y in x.
[tex]f(x)=y\\ 2^x+3=y\\ 2^x=y-3\\ log_{2}2^x=log_{2}(y-3)\\ x=log_{2}(y-3)\\ g(x)=log_{2}(x-3)\\ g(5)=log_{2}(5-3)=log_{2}(2)=1[/tex]