Question

Dau coroana. Ma interesewza doar punctul c).

Answer 1

obesrvi de la a) ca functai are un maxim pt 3-9x=0 adica x=1/3 (derivata fiind pozitiva inainte si negativa dupa1/3)

f(1/3)=(9si1/3)/√(1/9+3)=28/3:√(28/9)=28:√28=√28=2√7

de asemenea, mai obserc vi, facand limitele la + si- infinit ,eventual la pctul b), ca acestea sunt 1 , respectiv, -1

dci imaginea functiei este (-1;2√7]

nu prea usoara, ai o explicatie mai "pe scurt", ca de 5-:-8 puncte

Answer 2

$\it f'(x) =\dfrac{3-9x}{(x^2+3)\sqrt{x^2+3}};\ \ \ f'(x) =0 \Rightarrow 3-9x=0\Rightarrow x=\dfrac{1}{3} \\ \\ \\ f\left(\dfrac{1}{3}\right)= 2\sqrt{7} \\ \\ \\ Se \ constat\breve{a}\ c\breve{a} \ A\left(\dfrac{1}{3}, 2\sqrt7\right) \ este\ punct\ de \ maxim, \ deci\ f(x) \leq2\sqrt7\ \ \ (1)$

$\it \lim\limits_{x \to -\infty} f(x) = \lim\limits_{x \to -\infty} \dfrac{x+9}{\sqrt{x^2+3}} = \lim\limits_{x \to -\infty} \dfrac{x(1+\dfrac{9}{x})}{|x|\sqrt{1+\dfrac{3}{x^2}}} = \lim\limits_{x \to -\infty} \dfrac{x}{-x} =-1\Rightarrow \\ \\ \\ \Rightarrow f(x) >-1 \ \ \ (2) \\ \\ \\ (1),\ (2) \Rightarrow -1<f(x) \leq2\sqrt7 \Rightarrow Imf \in(-1,\ 2\sqrt7]$