Question

Va rog dau coroana exercitiul 24 ​

Answer 1

Explicație pas cu pas:

$f(x) = x ln(x)$

a) l'Hospital:

$lim_{\left \{ {{x \rightarrow 0 } \atop {x > 0}} \right.}(f(x)) = lim_{\left \{ {{x \rightarrow 0 } \atop {x > 0}} \right.}(x ln(x)) \\ = lim_{\left \{ {{x \rightarrow 0 } \atop {x > 0}} \right.}\left(\frac{ln(x)}{ \frac{1}{x} } \right) = lim_{\left \{ {{x \rightarrow 0 } \atop {x > 0}} \right.}\left(\frac{ \frac{1}{x} }{ - \frac{1}{ {x}^{2} } } \right) \\ = lim_{\left \{ {{x \rightarrow 0 } \atop {x > 0}} \right.}( - x) = 0$

b) prima derivată:

$f'(x) = \left(xln(x) \right)' = x'ln(x) + x\left(ln(x) \right)' \\ = ln(x) + x \cdot \frac{1}{x} = ln(x) + 1$

$f'(x) = 0 => ln(x) + 1 = 0 \\ ln(x) = - 1 = > x = {e}^{ - 1}$

$f\left( \frac{1}{e} \right) = \frac{ ln( {e}^{ - 1} ) }{e} = - \frac{1}{e} \\ = > punct \: de \: minim : \left( \frac{1}{e}; - \frac{1}{e}\right)$

intervale de monotonie:

$0 < x < \frac{1}{e} = > f(x) \: descrescatoare$

$\frac{1}{e} < x < + \infty = > f(x) \: crescatoare$

c) derivata a doua:

$f''(x) = (f'(x))' = (ln(x) + 1)' = \frac{1}{x} > 0, x \in (0; +\infty)$

=> f(x) este convexă