[tex]f(x)=\sqrt{x^{2}+4 x+5}-x-2[/tex]
a)
Vezi tabelul de derivate din atasament
[tex]f'(x)=\frac{2x+4}{2\sqrt{x^2+4x+5} }-1=\frac{x+2}{\sqrt{x^2+4x+5} } -1[/tex]
b)
Asimptota orizontala spre +∞
Calculam limita spre +∞
[tex]\lim_{x \to +\infty} \sqrt{x^2+4x+5} -x-2= \lim_{x \to +\infty} \sqrt{x^2+4x+5} -(x+2)[/tex]
Facem conjugata si obtinem
[tex]\lim_{x \to +\infty} \sqrt{x^2+4x+5} -(x+2)= \lim_{x \to +\infty} \frac{x^2+4x+5-(x^2+4x+4)}{\sqrt{x^2+4x+5}+x+2 } =\\\\= \lim_{x \to +\infty} \frac{1}{\sqrt{x^2+4x+5}+x+2 } =\frac{1}{\infty} =0[/tex]
c)
f'(x)=0
[tex](x+2)-\sqrt{x^2+4x+5} =0\\\\x+2=\sqrt{x^2+4x+5}\ \ |^2\\\\x^2+4x+4=x^2+4x+5\\\\4=5\ NU\\\\f'(x)\neq 0[/tex]
[tex]f'(0)=\frac{2}{\sqrt{5} } -1 < 0[/tex]⇒f este descrescatoare
[tex]\lim_{x \to -\infty} f(x)=+\infty\\\\ \lim_{x \to +\infty} f(x)=0[/tex]⇒Imaginea functiei f este (0,+∞)
Un alt exercitiu cu fuunctii gasesti aici: https://brainly.ro/tema/9918951
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