[tex]f(x)=\frac{x-1}{x+1}+ln(x+1)-lnx[/tex]
Facem monotonia functiei f
f'(x)=0
[tex]f'(x)=\frac{x+1-(x-1)}{(x+1)^2} +\frac{1}{x+1} -\frac{1}{x}\\\\f'(x)=\frac{2}{(x+1)^2} +\frac{x-x-1}{x(x+1)} \\\\f'(x)=\frac{2x-(x+1)}{x(x+1)^2} =\frac{x-1}{x(x+1)^2} \\\\f'(x)=0\\\\[/tex](Vezi tabelul de derivate din atasament)
x-1=0
x=1
Tabel semn
x 0 1 +∞
f'(x) - - - - - -0 + + + + + +
f(x) ↓ f(1) ↑
ln2
f(1)=0+ln2+ln1=ln2+0=ln2
ln2>0⇒ Gf nu intersecteaza axa Ox
Un alt exercitiu cu functii gasesti aici: https://brainly.ro/tema/2324458
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