Question

URGENT VA ROG!!!!!!!

Answer 1

f(x)=x-2lnx

$f'(x)=1-\frac{2}{x}=\frac{x-2}{x}$

(vezi tabelul de derivare din atasament)

Monotonia functiei f

Calculam f'(x)=0

$\frac{x-2}{x}=0$

x-2=0

x=2

Facem tabel semn

x         1          2               +∞

f'(x)     - - - - - 0 + + + + + +

f(x)         ↓    f(2)      ↑

                   2-2ln2

f(2)=2-2ln2 -punctul de extrem local

Convexitatea

Calculam f''(x)

$f''(x)=(1-\frac{2}{x})'=0-(-\frac{2}{x^2})=\frac{2}{x^2}$

Observam ca f''(x)>0⇒ f(x) este convexa pe [1,+∞)

f(e)=e-2lne=e-2

$\lim_{x \to e} \frac{f(x)-f(e)}{x^2-e^2}=\frac{0}{0}$

Aplicam L'Hospital (derivam numarator, derivam numitor)

$\lim_{x \to e} \frac{f'(x)-0}{2x}= \lim_{x \to e} \frac{\frac{x-2}{x} }{2x} = \lim_{x \to e} \frac{x-2}{2x^2}=\frac{e-2}{2e^2}$

sau

$\lim_{x \to e} \frac{f(x)-f(e)}{(x-e)(x+e)} = f'(e)\cdot \lim_{x \to e} \frac{1}{x+e} =\frac{e-2}{e}\cdot \frac{1}{2e}=\frac{e-2}{2e^2}$

Ne folosim de faptul ca:

$\lim_{x \to e} \frac{f(x)-f(e)}{x-e}=f'(e)$

Mai multe despre monotonia unei functii gasesti aici: https://brainly.ro/tema/978074

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