[tex]f(x)=x-\ln \left(2^{x}+1\right)$[/tex]
a)
Vezi tabelul de derivate din atasament
a)
[tex]f'(x)=1-\frac{2^xln2}{2^x+1}[/tex]
b)
Monotonia functiei f
[tex]0 < \frac{2^x}{2^x+1} < 1\ \ \ |\cdot ln2\\\\0 < \frac{2^xln2}{2^x+1} < ln2\ \ \ |\cdot (-1)\\\\0 > -\frac{2^xln2}{2^x+1} > ln2\ \ \ |+1\\\\1 > 1-\frac{2^xln2}{2^x+1} > 1+ln2[/tex]
f'(x)>0⇒ f este crescatoare
c)
Ecuatia asimptotei oblice
y=mx+n
[tex]m= \lim_{x \to -\infty} \frac{f(x)}{x} =\lim_{x \to -\infty} \frac{x-ln(2^x+1)}{x} =\lim_{x \to -\infty} \frac{x}{x} -\frac{ln(2^x+1)}{x} =1-0=1\\\\n=\lim_{x \to -\infty} f(x)-mx=\lim_{x \to -\infty} x-ln(2^x+1)-x=\lim_{x \to -\infty} -ln(2^x+1)=-ln1=0[/tex]
y=x
Un alt exercitiu cu functii gasesti aici: https://brainly.ro/tema/9928394
#BAC2022
#SPJ4
