[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]
