[tex]f(x)=\sqrt{\frac{x+1}{x-1}}[/tex]
a)
Vezi tabelul de derivate din atasament
[tex]f'(x)=\frac{\frac{1}{2\sqrt{x+1} }\sqrt{x-1}-\sqrt{x+1} \frac{1}{2\sqrt{x-1} } }{(\sqrt{x-1})^2 } =\frac{\frac{-2}{2\sqrt{x-1}\sqrt{x+1} } }{x-1} =-\frac{1}{(x-1)\sqrt{x^2-1} }[/tex]
b)
Ecuatia tangentei in punctul A(a,f(a))
y-f(a)=f'(a)(x-a)
In cazul nostru a=2
[tex]f(2)=\sqrt{3} \\\\f'(2)=-\frac{\sqrt{3} }{3}[/tex]
[tex]y-\sqrt{3} =-\frac{\sqrt{3} }{3}(x-\sqrt{3})\\\\\\y=-\frac{\sqrt{3} }{3}x+1+\sqrt{3}[/tex]
c)
[tex]\lim_{x \to 1} f(x)=\frac{\sqrt{2} }{0}=+\infty[/tex] ⇒ avem asimptota verticala , ecuatia dreptei este x=1
[tex]\lim_{x \to+ \infty} f(x)=\frac{\infty}{\infty}[/tex]
[tex]\lim_{x \to +\infty} \frac{\sqrt{x(1+\frac{1}{x}) }}{\sqrt{x(1-\frac{1}{x} )} } =1\\\\\frac{1}{x} \to0[/tex]
Dreapta de ecuatie x=1 este asimptota orizontala
Punctul de intersectie este (1,1)
Un alt exercitiu cu functii gasesti aici: https://brainly.ro/tema/9918934
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