Question

Ma interesează oricare 2 exerciții rezolvate de aici. ​

Answer 1

a.

$f(x)=x+\sqrt{x^2+1}$

$f'(x)=1+\frac{x}{\sqrt{x^2+1} }$

$\frac{x}{\sqrt{x^2}+1 }$ este o fractie subunitara

Deci daca x<0

$1+\frac{x}{\sqrt{x^2}+1 } > 0$

Daca x>0

$1+\frac{x}{\sqrt{x^2}+1 } > 0$

⇒ f este strict crescatoare

b.

$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} }$$(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)$

c.

$\lim_{x \to \infty} f(x)=\infty+\infty=\infty$ deci nu avem asimptota orizontala

calculam asimptota oblica

$m= \lim_{x \to \infty} \frac{x+\sqrt{x^2+1} }{x} = \lim_{x \to \infty}1+\frac{\sqrt{x^2(1+\frac{1}{x^2}) } }{x} = \lim_{x \to \infty} 1+\sqrt{1+\frac{1}{x^2} } =1+1=2\\\\\frac{1}{x^2}=\frac{1}{\infty} =0$

$n= \lim_{x \to \infty} x+\sqrt{x^2+1} -2x=\sqrt{x^2+1}-x=\infty-\infty=\ forma\ nedeterminata\\\\n= \lim_{x \to \infty} \frac{(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)}{\sqrt{x^2+1}+x} \\\\n= \lim_{x \to \infty} \frac{1}{\sqrt{x^2+1}+x} =\frac{1}{\infty} =0$

y=mx+n

y=2x