Explicație pas cu pas:
[tex]f(x) = \frac{{x}^{2} - 1}{{x}^{2} + 1} \\[/tex]
a)
[tex]f^{\prime}(x) = (\frac{{x}^{2} - 1}{{x}^{2} + 1})^{\prime} \\ = \frac{( {x}^{2} - 1)^{\prime}({x}^{2} + 1) - ({x}^{2} - 1)({x}^{2} + 1)^{\prime}}{ {( {x}^{2} + 1)}^{2} } \\ = \frac{2x\cdot( {x}^{2} + 1) - ( {x}^{2} - 1)\cdot2x}{ {( {x}^{2} + 1)}^{2}} \\ = \frac{2 {x}^{3} + 2x - 2 {x}^{3} + 2x}{ {( {x}^{2} + 1)}^{2} } = \frac{4x}{ {( {x}^{2} + 1)}^{2} }[/tex]
b)
[tex]lim_{x \rightarrow 1}\left(\frac{f(x)}{x - 1} \right) = lim_{x \rightarrow 1}\left(\frac{ \frac{{x}^{2} - 1}{{x}^{2} + 1}}{x - 1} \right) \\ = lim_{x \rightarrow 1}\left(\frac{(x - 1)(x + 1)}{ {( {x}^{2} + 1)}^{2}(x - 1) } \right) = lim_{x \rightarrow 1}\left(\frac{x + 1}{ {( {x}^{2} + 1)}^{2} } \right) \\ = \frac{1 + 1}{ {( {1}^{2} + 1)}^{2} } = \frac{2}{4} = \frac{1}{2}[/tex]
c)
[tex]f''(x) = \left(f'(x) \right) = \left(\frac{4x}{ {( {x}^{2} + 1)}^{2} } \right)' = 4\cdot\left(\frac{x}{ {( {x}^{2} + 1)}^{2} } \right)' \\= 4\cdot\frac{ x^{\prime}{( {x}^{2} + 1)}^{2} - x \left( {( {x}^{2} + 1)}^{2} \right)^{\prime}}{ ({( {x}^{2} + 1)}^{2})^{2} } \\= 4\cdot\frac{ {( {x}^{2} + 1)}^{2} - x\cdot2( {x}^{2} + 1) \left( { {x}^{2} + 1} \right)^{\prime}}{ {( {x}^{2} + 1)}^{4} } \\= 4\cdot\frac{ {( {x}^{2} + 1)}^{2} - x\cdot2( {x}^{2} + 1)\cdot2x}{ {( {x}^{2} + 1)}^{4} } \\= 4\cdot\frac{ {( {x}^{2} + 1)}( {x}^{2} + 1 - 4 {x}^{2} )}{ {( {x}^{2} + 1)}^{4} } \\= 4\cdot\frac{(1 - 3 {x}^{2} )}{ {( {x}^{2} + 1)}^{3} }[/tex]
[tex]f''(x) = 0 = > - 3 {x}^{2} + 1 = 0[/tex]
[tex]x_{1} = - \frac{ \sqrt{3} }{3} ; \: x_{2} = \frac{ \sqrt{3} }{3} \\ [/tex]
[tex]f''(x) \geqslant 0 = > x\in \left[ - \frac{ \sqrt{3} }{3} ; \frac{ \sqrt{3} }{3} \right][/tex]
[tex] = > f(x) \: este \: convexa \: pentru: \\ x\in \left[ - \frac{ \sqrt{3} }{3} ; \frac{ \sqrt{3} }{3} \right][/tex]
