a.
[tex]f(x)=x+\sqrt{x^2+1}[/tex]
[tex]f'(x)=1+\frac{x}{\sqrt{x^2+1} }[/tex]
[tex]\frac{x}{\sqrt{x^2}+1 }[/tex] este o fractie subunitara
Deci daca x<0
[tex]1+\frac{x}{\sqrt{x^2}+1 } > 0[/tex]
Daca x>0
[tex]1+\frac{x}{\sqrt{x^2}+1 } > 0[/tex]
⇒ f este strict crescatoare
b.
[tex]f''(x)=\frac{\sqrt{x^2+1} -x\times \frac{x}{\sqrt{x^2+1} } }{x^2+1} \\\\f''(x)=\frac{1}{(x^2+1)\sqrt{x^2+1} }[/tex][tex](x^2+1)f''(x)f(x)=(x^2+1)\times\frac{1}{(x^2+1)\sqrt{x^2+1} }\times(x+\sqrt{x^2+1)} =\frac{1}{\sqrt{x^2+1} }\times (x+\sqrt{x^2+1} =\frac{x}{\sqrt{x^2+1} } +1=f'(x)[/tex]
c.
[tex]\lim_{x \to \infty} f(x)=\infty+\infty=\infty[/tex] deci nu avem asimptota orizontala
calculam asimptota oblica
[tex]m= \lim_{x \to \infty} \frac{x+\sqrt{x^2+1} }{x} = \lim_{x \to \infty}1+\frac{\sqrt{x^2(1+\frac{1}{x^2}) } }{x} = \lim_{x \to \infty} 1+\sqrt{1+\frac{1}{x^2} } =1+1=2\\\\\frac{1}{x^2}=\frac{1}{\infty} =0[/tex]
[tex]n= \lim_{x \to \infty} x+\sqrt{x^2+1} -2x=\sqrt{x^2+1}-x=\infty-\infty=\ forma\ nedeterminata\\\\n= \lim_{x \to \infty} \frac{(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)}{\sqrt{x^2+1}+x} \\\\n= \lim_{x \to \infty} \frac{1}{\sqrt{x^2+1}+x} =\frac{1}{\infty} =0[/tex]
y=mx+n
y=2x
