Question

Multimea valorilor functiei f:R->R; f(x)=$<var>\sqrt{x^{2}+x+1}-\sqrt{x^{2}+x-1}</var>$

Answer 1

[tex]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}}} [/tex]
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 :$(0;\sqrt{2})$.