Explicație pas cu pas:
[tex]f(x) = x ln(x) [/tex]
a) l'Hospital:
[tex]lim_{\left \{ {{x \rightarrow 0 } \atop {x > 0}} \right.}(f(x)) = lim_{\left \{ {{x \rightarrow 0 } \atop {x > 0}} \right.}(x ln(x)) \\ = lim_{\left \{ {{x \rightarrow 0 } \atop {x > 0}} \right.}\left(\frac{ln(x)}{ \frac{1}{x} } \right) = lim_{\left \{ {{x \rightarrow 0 } \atop {x > 0}} \right.}\left(\frac{ \frac{1}{x} }{ - \frac{1}{ {x}^{2} } } \right) \\ = lim_{\left \{ {{x \rightarrow 0 } \atop {x > 0}} \right.}( - x) = 0[/tex]
b) prima derivată:
[tex]f'(x) = \left(xln(x) \right)' = x'ln(x) + x\left(ln(x) \right)' \\ = ln(x) + x \cdot \frac{1}{x} = ln(x) + 1[/tex]
[tex]f'(x) = 0 => ln(x) + 1 = 0 \\ ln(x) = - 1 = > x = {e}^{ - 1} [/tex]
[tex]f\left( \frac{1}{e} \right) = \frac{ ln( {e}^{ - 1} ) }{e} = - \frac{1}{e} \\ = > punct \: de \: minim : \left( \frac{1}{e}; - \frac{1}{e}\right)[/tex]
intervale de monotonie:
[tex]0 < x < \frac{1}{e} = > f(x) \: descrescatoare \\ [/tex]
[tex]\frac{1}{e} < x < + \infty = > f(x) \: crescatoare \\ [/tex]
c) derivata a doua:
[tex]f''(x) = (f'(x))' = (ln(x) + 1)' = \frac{1}{x} > 0, x \in (0; +\infty)\\ [/tex]
=> f(x) este convexă
