Răspuns :
[tex]f(x)=\frac{1}{(x-1)^{2}}-\frac{1}{x^{2}}[/tex]
a)
Ne folosim de formula:
[tex](\frac{f}{g} )'=\frac{f'\times g-f\times g'}{g^2}[/tex]
Calculam f'(x)
[tex]f'(x)=(\frac{1}{(x-1)^{2}})'-(\frac{1}{x^{2}})'=\frac{1'\times(x-1)^2-((x-1)^2)'}{(x-1)^4} -\frac{1'\times x^2-(x^2)'}{x^4}[/tex]
[tex]f'(x)=\frac{-2(x-1)}{(x-1)^4} -\frac{-2x}{x^4} =\frac{-2}{(x-1)^3} +\frac{2}{x^3}[/tex]
Aducem la acelasi numitor, prima fractie o amplificam cu x³ si a doua cu (x-1)³ si vom avea aceeasi linie de fractie
[tex]f'(x)=\frac{-2x^3+2(x-1)^3}{x^3(x-1)^3} =\frac{-2x^3+2x^3-6x^2+6x-2}{x^3(x-1)^3} =\frac{-6x^2+6x-2}{x^3(x-1)^3} =\frac{-2(3x^2-3x+1)}{x^3(x-1)^3}[/tex]
b) Ecuatia unei drepte care trece printr-un punct A
[tex]y-y_A=m(x-x_A)[/tex], unde m=panta dreptei
Daca 2 drepte sunt paralele atunci pantele vor fi egale
Panta dreptei care este paralela cu tangenta la graficul functiei f in punctul de abscisa x=2 este f'(2)
[tex]f'(2)=\frac{-2(3\times 2^2-3\times 2+1)}{2^3(2-1)^3}=\frac{-14}{8} =-\frac{7}{4}[/tex]
Ecuatia dreptei va fi:
[tex]y-3=-\frac{7}{4} (x-0)\\\\y-3=-\frac{7x}{4} \\\\y=-\frac{7x}{4}+3[/tex]
c) Ne folosim de limita remarcabila
[tex]\lim_{x \to \infty} (1+\frac{1}{x} )^x=e[/tex]
Calculam f(2)+f(3)+...+f(n)
[tex]f(2)+f(3)+...+f(n)=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9} +...+\frac{1}{(n-1)^2} -\frac{1}{n^2}[/tex]
Observam ca se reduc termenii si ne ramane:
[tex]f(2)+f(3)+...+f(n)=1-\frac{1}{n^2}[/tex]
Calculam limita si obtinem:
[tex]\lim_{n \to \infty} (1-\frac{1}{n^2} )^{n^2}= \lim_{n \to \infty} [(1+(-\frac{1}{n^2} )^-^{n^2}]^{-1}=e^{-1}=\frac{1}{e}[/tex]
Un exercitiu similar de bac gasesti aici: https://brainly.ro/tema/3635705
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