Răspuns :
[tex]f(x)=(x+1) e^{-x}[/tex]
a)
Derivam functia f prin formula (fg)'=f'g+fg'
f'(x)=(x+1)'e⁻ˣ +(x+1)(e⁻ˣ)'=e⁻ˣ-(x+1)e⁻ˣ=e⁻ˣ(1-x-1)=-xe⁻ˣ
b)
f(n)=(n+1)e⁻ⁿ
(f(n))ⁿ=[(n+1)e⁻ⁿ]ⁿ
f(n+1)=(n+2)e⁻ⁿ⁻¹
(f(n+1))ⁿ=[(n+2)e⁻ⁿ⁻¹]ⁿ
[tex]\lim_{n \to \infty} \frac{[(n+1)e^{-n}]^n}{e^n[(n+2)e^{-n-1}]^n} = \lim_{n \to \infty} (\frac{(n+1)e^{-n}}{e(n+2)e^{-n-1}} )^n= \lim_{n \to \infty} (\frac{(n+1)e^{-n}}{(n+2)e^{-n}} )^n[/tex]
Se simplifica e⁻ⁿ si ne ramane:
[tex]\lim_{n \to \infty}(\frac{n+1}{n+2})^n=1^{\infty}[/tex]
Avem o limita remarcabila in care vom face urmatorul artificiu de calcul:
[tex]\lim_{n \to \infty}(\frac{n+1}{n+2})^n=\lim_{n \to \infty}(\frac{n+2-1}{n+2})^n= \lim_{n \to \infty}[ (1+\frac{-1}{n+2})^{-(n+2)} ]^{\frac{-n}{n+2}}=\\\\=e ^{ \lim_{n \to \infty} \frac{-n}{n+2} }=e^{-1} =\frac{1}{e}[/tex]
c)
f'(x)=0
-xe⁻ˣ=0
x=0
Facem tabel semn
x -∞ 0 +∞
f'(x) + + + + +0 - - - - -
f(x) ↑ f(0) ↓
1
f(0)=(0+1)e⁰=1
f este crescatoare pe (-∞,0) si descrescatoare pe (0,+∞) ⇒ f are doua solutii reale distincte, m∈(0,1)
Un exercitiu similar cu limite gasesti aici: https://brainly.ro/tema/4614641
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