Question

abscisele punctelor in care graficele functiilor f,g:R-->R

f(x)=x^6 g(x)=2x^5-2x-1 sunt tangente

Answer 1

   
Daca graficele a doua functii f(x) si g(x) sunt tangente intr-un punct, atunci abscisa acelui punct este radacina dubla a ecuatiei f(x)=g(x).

Daca graficele au mai multe uncte in care sunt tangente atunvci abscisa fiecarui punct de tangenta va fi radacina dubla. 

Daca o radacina reala (sau mai multe) nu este dubla atunci exista un punct (sau mai multe) in care graficele se intersecteaza.

[tex]\displaystyle\\ \text{Rezolvam ecuatia }~f(x) = g(x)\\ f(x) = x^6\\ g(x)=2x^5-2x-1\\\\ x^6 = 2x^5-2x-1\\\\ x^6 - 2x^5+2x+1=0\\ \text{Prelucram ecuatia astfel:}\\ -2x^5 = - x^5 - x^5 \\ +2x = + x + x\\ \text{Adaugam puterile lipsa:} \\ 0x^4 = -x^4 + x^4\\ 0x^3 = +x^3 -x^3\\ 0x^2 = + x^2 -x^2\\ \text{Ecuatia: }\\ x^6 - 2x^5+2x+1=0\\ \text{devine: } \\\\ x^6 - x^5 -x^4 - x^5 + x^4 +x^3 -x^3+ x^2+ x -x^2+ x + 1 =0\\[/tex]

[tex]\displaystyle\\ \text{Rezolvare:}\\ x^6 - x^5 -x^4 - x^5 + x^4 +x^3 -x^3+ x^2+ x -x^2+ x + 1 =0\\\\ (x^6 - x^5 -x^4) - (x^5 - x^4 -x^3) -(x^3- x^2- x) -(x^2- x - 1) =0\\\\ x^4(x^2 - x -1) - x^3(x^2 - x -1) - x(x^2- x- 1) -1(x^2- x - 1) =0\\\\ (x^2 - x -1)(x^4- x^3- x-1)=0\\\\ (x^2 - x -1)[x^4- x^3 - x^2 + x^2 - x -1]=0\\\\ (x^2 - x -1)[(x^4- x^3 - x^2) + (x^2 - x -1)]=0\\\\ (x^2 - x -1)[ x^2(x^2- x - 1) + 1(x^2 - x -1)]=0\\\\ (x^2 - x -1)[ (x^2- x-1)(x^2+1)]=0\\\\ (x^2 - x -1)(x^2 - x -1)(x^2+1)=0 [/tex]


[tex]\displaystyle\\ (x^2-x-1)(x^2-x-1)(x^2+1)=0\\ \text{Ecuatia 1: }~x^2-x-1=0\\ \text{Ecuatia 2: }~x^2-x-1 = 0\\ \text{Ecuatia 3: }~x^2+1=0\\ \text{Avem 2 ecuatii identice.}\\ \text{Rezolvam ecuatia 1: }~x^2-x-1=0\\\\ x_{12}=\frac{1\pm\sqrt{1+4}}{2}=\frac{1\pm\sqrt{5}}{2}\\\\ x_1=\frac{1+\sqrt{5}}{2}\\\\ x_2=\frac{1-\sqrt{5}}{2}\\\\ \text{Rezolvam ecuatia 2: }~x^2-x-1=0\\\\ x_{34}=\frac{1\pm\sqrt{1+4}}{2}=\frac{1\pm\sqrt{5}}{2}\\\\ x_3=\frac{1+\sqrt{5}}{2}\\\\ x_4=\frac{1-\sqrt{5}}{2}\\\\ [/tex]

[tex]\displaystyle\\ \text{Rezolvam ecuatia 3: }~x^2+1=0\\ x^2 = -1\\ x_{56} = \pm \sqrt{-1} =\pm i\\ x_5 = i \notin R\\ x_6 = -i \notin R\\\\ \text{Observam ca:}\\\\ x_1 = x_3 =\frac{1+\sqrt{5}}{2} \\\\ x_2 = x_4 =\frac{1-\sqrt{5}}{2} \\\\ \text{Rezulta ca graficele sunt tangente in 2 puncte ale caror abscise sunt:}\\\\ \boxed{\frac{1+\sqrt{5}}{2}~~\text{respectiv}~~\frac{1-\sqrt{5}}{2} }[/tex]