Explicație pas cu pas:
[tex]f(x) = x^{2}e^{x}[/tex]
a) limita:
[tex]\lim _{x \rightarrow - \infty } f(x) = \lim _{x \rightarrow - \infty } \left(x^{2}e^{x} \right) \\ = \lim _{x \rightarrow - \infty } \left( \frac{x^{2}}{ \frac{1}{e^{x}} }\right) = \lim _{x \rightarrow - \infty } \frac{(x^{2})'}{\left(\frac{1}{e^{x}} \right)'} \\ = \lim _{x \rightarrow - \infty } \left( \frac{2x}{ - \frac{1}{e^{x}} }\right) = \lim _{x \rightarrow - \infty } \frac{(2x)'}{\left( - \frac{1}{e^{x}} \right)'} \\ = \lim _{x \rightarrow - \infty } \left( \frac{2}{ \frac{1}{e^{x}} }\right) = 2\lim _{x \rightarrow - \infty }e^{x} = 2 \cdot 0 = 0[/tex]
b) intervale de monotonie:
prima derivată
[tex]f'(x) = \left(x^{2}e^{x} \right)' = (x^{2})'e^{x} + x^{2}(e^{x})' \\ = 2xe^{x} + x^{2}e^{x} = xe^{x}( x + 2)[/tex]
[tex]f'(x) = 0 = > xe^{x}( x + 2) = 0[/tex]
→
[tex]x = 0[/tex]
[tex]f(0) = 0 = > minim \: \left(0 ; 0\right)[/tex]
→
[tex]x + 2 = 0 = > x = - 2[/tex]
[tex]f( - 2) = 4 {e}^{ - 2} = > maxim \: \left( - 2 ; 4 {e}^{ - 2}\right) \\ [/tex]
intervale de monotonie:
[tex] - \infty < x < - 2 = > f(x) \: crescatoare \\ [/tex]
[tex] - 2 < x < 0= > f(x) \: descrescatoare \\ [/tex]
[tex]0 < x < + \infty = > f(x) \: crescatoare[/tex]
c) derivata a doua:
[tex]f''(x) = \left(f'(x) \right)' = \left(2xe^{x} + x^{2}e^{x} \right)'
\\ = \left(2xe^{x}\right)' + \left(x^{2}e^{x} \right)' = 2e^{x} + 2xe^{x} + 2xe^{x} + x^{2}e^{x} \\ = x^{2}e^{x} + 4xe^{x} + 2e^{x} = e^{x}(x^{2} + 4x + 2)[/tex]
→
[tex]f''(x) = 0 => e^{x}(x^{2} + 4x + 2) = 0[/tex]
[tex]x^{2} + 4x + 2 = 0[/tex]
[tex]x_{1} = -2 - \sqrt{2} ; \: x_{2} = -2 + \sqrt{2} [/tex]
[tex]f''(x) \leqslant 0 => x \in [-2 - \sqrt{2} ; -2 + \sqrt{2}] \\ [/tex]
=> f(x) este concavă pe intervalul:
[tex]x \in [-2 - \sqrt{2} ; -2 + \sqrt{2}][/tex]
