Question

Se consideră funcţia $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=\frac{x^{2}+4 x+4}{e^{x}}$.

$5 p$ a) Arătaţi că $f^{\prime}(x)=\frac{-x(x+2)}{e^{x}}, x \in \mathbb{R}$.

$5 \mathbf{p}$ b) Determinaţi ecuaţia asimptotei orizontale spre $+\infty$ la graficul funcţiei $f$.

$5 p$ c) Demonstrați că $\lim _{n \rightarrow+\infty}(g(1)+g(2)+\ldots+g(n))=\frac{1}{e-1}$, unde $g:(0,+\infty) \rightarrow \mathbb{R}, g(x)=\frac{f(x)}{(x+2)^{2}}$.

Answer 1

$f(x)=\frac{x^{2}+4 x+4}{e^{x}}$

a)

Folosim formula de derivare a unei fractii:

$(\frac{f}{g} )'=\frac{f'g-fg'}{g^2}$

$f'(x)=(\frac{x^{2}+4 x+4}{e^{x}})'=\frac{(2x+4)e^x-(x^2+4x+4)e^x}{e^{2x}} =\frac{e^x(2x+4-x^2-4x-4)}{e^{2x}} =\frac{-x^2-2x}{e^x} =\frac{-x(x+2)}{e^x}$

b)

Trebuie sa calculam limita spre +∞ din functia f(x)

$\lim_{x \to \infty} \frac{x^{2}+4 x+4}{e^{x}}=\frac{\infty}{\infty} \ forma\ nedeterminata$

Aplicam L'Hospital, derivam numarator, derivam numitor si vom obtine:

$\lim_{x \to \infty} \frac{(x^{2}+4 x+4)'}{(e^{x})'}=\lim_{x \to \infty} \frac{2x+4}{e^{x}}=\frac{\infty}{\infty} \ forma\ nedeterminata$

Aplicam iar L'Hospital si vom obtine:

$\lim_{x \to \infty} \frac{(2x+4)'}{(e^{x}})'=\lim_{x \to \infty} \frac{2}{e^{x}}=\frac{2}{\infty} =0$

Ecuatia asimptotei orizontale este y=0

c)

$g(x)=\frac{\frac{x^2+4x+4}{e^x} }{(x+2)^2} =\frac{\frac{(x+2)^2}{e^x} }{(x+2)^2} =\frac{1}{e^x}\\\\g(x)=\frac{1}{e^x}$

Calculam g(1)+g(2)+...+g(n)

$g(1)+g(2)+...+g(n)=\frac{1}{e} +\frac{1}{e^2} +...+\frac{1}{e^n}$

Avem suma unei progresii geometrice care are formula:

$S_n=b_1\frac{q^n-1}{q-1}$, unde $b_1=primul\ termen=\frac{1}{e} \\\\r=ratia=\frac{1}{e}$

$g(1)+g(2)+...+g(n)=\frac{1}{e} \times \frac{(\frac{1}{e})^n-1 }{\frac{1}{e}-1 }$

$\lim_{n \to \infty} \frac{1}{e} \times \frac{(\frac{1}{e})^n-1 }{\frac{1}{e}-1 } =-\frac{1}{e}\times \frac{1}{\frac{1-e}{e} } =\frac{1}{e-1}$

Nota:

$(\frac{1}{e} )^n,\ n- > +\infty\\\\(\frac{1}{e} )^\infty=0,\ deoarece\ este fractie\ subunitara$

Un exercitiu similar de bac gasesti aici: https://brainly.ro/tema/5855250

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