Răspunsuri

2014-07-16T16:20:46+03:00
f(x)=\sqrt{x^2+x+1}-\sqrt{x^2+x-1}=\frac{x^2+x+1-x^2-x+1}{\sqrt{x^2+x+1}+\sqrt{x^2+x-1}}=\\
= \frac{2}{\sqrt{x^2+x+1}+\sqrt{x^2+x-1}} >0=>f(x)>0\\
f(x)=\frac{2}{\sqrt{(x+\frac{1}{2})^+\frac{3}{4}}+\sqrt{((x+\frac{1}{2})^2-\frac{5}{4}}}
Acest raport ia valoarea maxima daca numitorul e cat mai mic. Al doilea radical de jos ia valoarea cea mai mica 0 pentru (x+\frac{1}{2})^2=\frac{5}{4}=>x+\frac{1}{2}=\pm\frac{\sqrt{5}}{2}
Inlocuind x+\frac{1}{2}=\pm\frac{\sqrt{5}}{2} in primul radical se obtine
\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}=\sqrt{\frac{5}{4}+\frac{3}{4}}=\sqrt{2}
0<f(x)<\frac{2}{\sqrt{2}}
Multimea valorilor functiei f este intervalul :(0;\sqrt{2}).