Question

Fie functia f:R→R, f(x)= $\frac{2x}{1+x^{2}}$
Determinati numarul punctelor de extrem ale functiei f.

Answer 1

[tex][ \frac{2x}{x^{2} + 1} ] ' = 2 \frac{(x)'(x^2+1) - (x)(x^2+1)' }{(x^2+1)^2}} = \frac{2(x^2+1)-2(x+0)(x+1)} {(x^2+1)^2} [/tex]  = $\frac {2(1-x^2)}{(x^2+1)^2} = -\frac{2(x^2-1)}{x^2+1}$

x = + / - 1 

f(1) = 1
f(-1)  = -1  



x __║-∞_____-1_______1_______+∞
-----------------------------------------------
f'(x)║↓↓↓↓↓0↑↑↑↑↑↑0↓↓↓↓↓
------------------------------------------------
f(x)║---------(-1)+++++++1-----------

Dupa cum putem observa din tabel, punctele de extrem sunt:

x= 1 Punct maxim (++++0-------)
x= -1 Punct minim (-----0+++++)