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Data publicarii: 22 Februarie, 2014

EXERCITIUL 2

Suport teoretic:

Derivate de ordin superior,descompuneri in factori,fractii simple.

Enunt:

Se da functia f:D - > R, definita prin legea

f(x)=\frac{x^2-6x-1}{x^3-2x^2-5x+6}.f(x)=\frac{x^2-6x-1}{x^3-2x^2-5x+6}.

1) Sa se determine domeniul sau maxim de definitie,  

D\subset{\mathbb{R}}D\subset{\mathbb{R}}

2) Sa se calculeze 

f^{(2014)}(0).f^{(2014)}(0).

Raspuns:

1)  D = R\{-2;1;3}.

2)\;f^{(2014)}(0)=2)\;f^{(2014)}(0)= (2014!)\cdot{[\frac{1!}{{2}^{2015}}}-1-\frac{1!}{{3}^{2015}}].(2014!)\cdot{[\frac{1!}{{2}^{2015}}}-1-\frac{1!}{{3}^{2015}}].

Rezolvare:

1) Trebuie eliminate radacinile reale ale ecuatiei:

x³ - 2x² - 5x + 6 = 0.

Se observa ca suma coeficientilor este egala cu zero, deci x = 1 este radacina

(aceasta ar putea fi identificata si prin testarea divizorilor termenului liber 9).

Apoi, ecuatia se poate rescrie astfel incat sa putem scoate ca factor comun binomul

(x - 1):

x³ - x² -  x² + x - 6x + 6 = 0 < = > x²(x - 1) - x(x - 1) - 6(x - 1 ) = 0 < =>

< = > (x - 1 )(x² - x - 6) = 0 < = > (x - 1)( x² - 3x + 2x - 6) = 0 <=>

< = > (x-1)[x(x-3) + 2(x-3)] = 0 < = >(x-1)(x-3)(x+2)= 0.

Rezulta ca D = R\{-2;1;3}.

2) Evident, calculul derivatei de ordinul 2014 ar fi extrem de laborios daca s-ar efectua

pe functia data in forma prezentata in enunt! Vom descompune, mai intai, in fractii

simple astfel:

f(x)=\frac{x^2-6x-1}{x^3-2x^2-5x+6}=\frac{A}{x+2}+\frac{B}{x-1}+\frac{C}{x-3}=\frac{A(x-1)(x-3)+B(x+2)(x-3)+C(x+2)(x-1)}{x^3-2x^2-5x+6}.f(x)=\frac{x^2-6x-1}{x^3-2x^2-5x+6}=\frac{A}{x+2}+\frac{B}{x-1}+\frac{C}{x-3}=\frac{A(x-1)(x-3)+B(x+2)(x-3)+C(x+2)(x-1)}{x^3-2x^2-5x+6}.

Cum egalitatea are loc pentru orice xЄD, se identifica numaratorii primei si ultimei fractii,

se obtine un sistem de ecuatii liniare cu necunoscutele A, B si C, dupa care:

f(x)=\frac{x^2-6x-1}{x^3-2x^2-5x+6}=\frac{1}{x+2}+\frac{1}{x-1}-\frac{1}{x-3},\;deci:f(x)=\frac{x^2-6x-1}{x^3-2x^2-5x+6}=\frac{1}{x+2}+\frac{1}{x-1}-\frac{1}{x-3},\;deci:

(f(x))^{(2014)}=(\frac{1}{x+2})^{(2014)}+(\frac{1}{x-1})^{(2014)}-(\frac{1}{x-3})^{(2014)}.(f(x))^{(2014)}=(\frac{1}{x+2})^{(2014)}+(\frac{1}{x-1})^{(2014)}-(\frac{1}{x-3})^{(2014)}.

Folosind o formula cunoscuta, anume  

{(\frac{1}{x-a})}^{(n)}={(-1)^{n}}\cdot{\frac{n!}{(x-a)^{n+1}}},\forall{a}\in{\mathbb{R}},\forall{x}\in{\mathbb{R^*}\setminus\{a\}},\;rezulta:{(\frac{1}{x-a})}^{(n)}={(-1)^{n}}\cdot{\frac{n!}{(x-a)^{n+1}}},\forall{a}\in{\mathbb{R}},\forall{x}\in{\mathbb{R^*}\setminus\{a\}},\;rezulta:  

(f(x))^{(2014)}=(-1)^{(2014)}\cdot{\frac{2014!}{{(x+2)}^{2015}}}+(-1)^{(2014)}\cdot{\frac{2014!}{{(x-1)}^{2015}}}+(-1)^{(2014)}\cdot{\frac{2014!}{{(x-3)}^{2015}}},\;deci:(f(x))^{(2014)}=(-1)^{(2014)}\cdot{\frac{2014!}{{(x+2)}^{2015}}}+(-1)^{(2014)}\cdot{\frac{2014!}{{(x-1)}^{2015}}}+(-1)^{(2014)}\cdot{\frac{2014!}{{(x-3)}^{2015}}},\;deci:

(f(x))^{(2014)}=(2014!)\cdot{[\frac{1!}{{(x+2)}^{2015}}}+\frac{1!}{{(x-1)}^{2015}}+\frac{1!}{{(x-3)}^{2015}}]\;si,\;in\;sfarsit:(f(x))^{(2014)}=(2014!)\cdot{[\frac{1!}{{(x+2)}^{2015}}}+\frac{1!}{{(x-1)}^{2015}}+\frac{1!}{{(x-3)}^{2015}}]\;si,\;in\;sfarsit:

f^{(2014)}(0)=(2014!)\cdot{[\frac{1!}{{(2)}^{2015}}}+\frac{1!}{{(-1)}^{2015}}+\frac{1!}{{(-3)}^{2015}}]=(2014!)\cdot{[\frac{1!}{{2}^{2015}}}-1-\frac{1!}{{3}^{2015}}].f^{(2014)}(0)=(2014!)\cdot{[\frac{1!}{{(2)}^{2015}}}+\frac{1!}{{(-1)}^{2015}}+\frac{1!}{{(-3)}^{2015}}]=(2014!)\cdot{[\frac{1!}{{2}^{2015}}}-1-\frac{1!}{{3}^{2015}}].


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