[tex]f(x)=\frac{x^{2}-1}{x^{2}+1}[/tex]
a)
Vezi tabel de derivate (in atasament)
[tex]f'(x)=\frac{2x(x^2+1)-2x(x^2-1)}{(x^2+1)^2} =\frac{2x^3+2x-2x^3+2x}{(x^2+1)^2}=\frac{4x}{(x^2+1)^2}[/tex]
b)
[tex]\lim_{x \to 1} \frac{\frac{(x-1)(x+1)}{x^2+1} }{x-1}= \lim_{n \to 1} \frac{x+1}{x^2+1}=1[/tex]
c)
Functie convexa:
Calculam derivata de ordin 2
[tex]f''(x)=(\frac{4x}{(x^2+1)^2})' =\frac{4(x^2+1)^2-8x^2(x^2+1)}{(x^2+1)^4} =\frac{(x^2+1)(4x^2+4-16x^2)}{(x^2+1)^4}=\frac{(4x^2+4-16x^2)}{(x^2+1)^3}=\\\\f''(x)=\frac{(4-12x^2)}{(x^2+1)^3}[/tex]
f''(x)=0
4-12x²=0
12x²=4
[tex]x^2=\frac{1}{3}\\\\ x_1=-\frac{\sqrt{3} }{3} \\\\x_2=\frac{\sqrt{3} }{3}[/tex]
Tabel semn
x -∞ [tex]-\frac{\sqrt{3} }{3}[/tex] [tex]\frac{\sqrt{3} }{3}[/tex] +∞
f''(x) - - - - - - - -0 + + + + 0 - - - - -
f(x) ↓ [tex]f(-\frac{\sqrt{3} }{3} )[/tex] ↑ [tex]f(\frac{\sqrt{3} }{3} )[/tex] ↓
f(x) este convexa pe intervalul [tex][-\frac{\sqrt{3} }{3} ,\frac{\sqrt{3} }{3} ][/tex]
Un alt exercitiu cu functii gasesti aici: https://brainly.ro/tema/9835783
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