Question

Ajutati la matem va rog

Answer 1

   
[tex]1) \displaystyle \\ f(x)= \frac{x^2}{x^2+1} \\ \text{Derivam functia folosind formula: } \\ \left(\frac{u}{v} \right)' = \frac{u'v-uv'}{v^2} \\ \\ f'(x) = \frac{2x(x^2+1)-x^2(2x+0) }{x^4+2x^2+1} =\frac{2x^3+2x-2x^3 }{x^4+2x^2+1} = \\ \\ =\frac{2x }{x^4+2x^2+1} \\ \\ 2x=0 \\ x=0 \\ \text{Functia are un minim in punctul: } ~~~O(0, ~0)[/tex]


[tex]2) \displaystyle \\ f(x)= \frac{2x}{x^2+9} \\ \text{Derivam functia folosind formula: } \\ \left(\frac{u}{v} \right)' = \frac{u'v-uv'}{v^2} \\ \\ f'(x) = \frac{2(x^2+9)-2x(2x+0) }{x^4+18x^2+81} = \frac{2x^2+18-4x^2}{x^4+18x^2+81} =\frac{-2x^2+18}{x^4+18x^2+81} \\ \\ -2x^2+18 =0 \\ x^2 = 9 \\ x_{12} = \pm \sqrt{9} \\ x_1 = 3 \\ x_2 = -3 [/tex]

[tex]\displaystyle f(3) = \frac{2 \cdot 3}{3^2 +9}= \frac{6}{18}= \frac{1}{3} \\ \Longrightarrow~~~Punctul~ de ~maxim~~A(3; ~ \frac{1}{3}) \\ \\ f(-3) = \frac{2 \cdot (-3)}{(-3^2) +9}= \frac{-6}{18}= \frac{-1}{3} \\ \Longrightarrow~~~Punctul~ de ~minim~~B(-3; ~ -\frac{1}{3}) \\ \\ \text{Voi atasa graficele celor 2 functii}[/tex]