Question

exercitiul 39 va rog dau coroana​

Answer 1

Explicație pas cu pas:

$f(x) = x^{2}e^{x}$

a) limita:

$\lim _{x \rightarrow - \infty } f(x) = \lim _{x \rightarrow - \infty } \left(x^{2}e^{x} \right) \\ = \lim _{x \rightarrow - \infty } \left( \frac{x^{2}}{ \frac{1}{e^{x}} }\right) = \lim _{x \rightarrow - \infty } \frac{(x^{2})'}{\left(\frac{1}{e^{x}} \right)'} \\ = \lim _{x \rightarrow - \infty } \left( \frac{2x}{ - \frac{1}{e^{x}} }\right) = \lim _{x \rightarrow - \infty } \frac{(2x)'}{\left( - \frac{1}{e^{x}} \right)'} \\ = \lim _{x \rightarrow - \infty } \left( \frac{2}{ \frac{1}{e^{x}} }\right) = 2\lim _{x \rightarrow - \infty }e^{x} = 2 \cdot 0 = 0$

b) intervale de monotonie:

prima derivată

$f'(x) = \left(x^{2}e^{x} \right)' = (x^{2})'e^{x} + x^{2}(e^{x})' \\ = 2xe^{x} + x^{2}e^{x} = xe^{x}( x + 2)$

$f'(x) = 0 = > xe^{x}( x + 2) = 0$

→

$x = 0$

$f(0) = 0 = > minim \: \left(0 ; 0\right)$

→

$x + 2 = 0 = > x = - 2$

$f( - 2) = 4 {e}^{ - 2} = > maxim \: \left( - 2 ; 4 {e}^{ - 2}\right)$

intervale de monotonie:

$- \infty < x < - 2 = > f(x) \: crescatoare$

$- 2 < x < 0= > f(x) \: descrescatoare$

$0 < x < + \infty = > f(x) \: crescatoare$

c) derivata a doua:

$f''(x) = \left(f'(x) \right)' = \left(2xe^{x} + x^{2}e^{x} \right)'</p><p> \\ = \left(2xe^{x}\right)' + \left(x^{2}e^{x} \right)' = 2e^{x} + 2xe^{x} + 2xe^{x} + x^{2}e^{x} \\ = x^{2}e^{x} + 4xe^{x} + 2e^{x} = e^{x}(x^{2} + 4x + 2)$

→

$f''(x) = 0 => e^{x}(x^{2} + 4x + 2) = 0$

$x^{2} + 4x + 2 = 0$

$x_{1} = -2 - \sqrt{2} ; \: x_{2} = -2 + \sqrt{2}$

$f''(x) \leqslant 0 => x \in [-2 - \sqrt{2} ; -2 + \sqrt{2}]$

=> f(x) este concavă pe intervalul:

$x \in [-2 - \sqrt{2} ; -2 + \sqrt{2}]$