Question

The maximum and minimum value of of $\frac{x^2+x+1}{x^2-x+1}$ is

• A. $2,1$
• B. $3,\frac{1}{3}$
• C. $1,0$
• D. $3,1$

Answer 1

The correct option is B $3,\frac{1}{3}$

Let $f\left(x\right)=\frac{x^2+x+1}{x^2-x+1}$
$f^{\prime }\left(x\right)=0$ implies
$(x^2-x+1)(2x+1)-(2x-1)(x^2+x+1)=0$
$2x^3-2x^2+2x+x^2-x+1=2x^3+2x^2+2x-x^2-x-1$
$-2x^2+x^2+1=2x^2-x^2-1$
$4x^2-2x^2-2=0$
$2x^2-2=0$
$\therefore x=\pm 1$
Hence, $f\left(1\right)$ $=\frac{3}{3-2}=3$
And
$f(-1)$ $=\frac{1-1+1}{1+1+1}$
$=\frac{1}{3}$.