Question

The range of the function \[ f(x)=\frac{x^2+x+1}{x^2-x+1} \] is

• A. \(\left[\frac{1}{3},3\right]\)
• B. \(\left[\frac{1}{2},2\right]\)
• C. \(\left[-\frac{1}{2},-\frac{1}{4}\right]\)
• D. \(\left[-\frac{1}{2},2\right]\)

Hint:

To find the range of a rational function, put \(y=f(x)\), form a quadratic equation in \(x\), and use the condition discriminant \(\geq 0\).

Answer 1

The Correct Option is A

Step 1: Let the function value be \(y\).
Let \[ y=\frac{x^2+x+1}{x^2-x+1} \] Since \[ x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\gt 0, \] the denominator is always positive.

Step 2: Convert into a quadratic equation in \(x\).
\[ y(x^2-x+1)=x^2+x+1 \] \[ yx^2-yx+y=x^2+x+1 \] Bring all terms to one side: \[ (y-1)x^2+(-y-1)x+(y-1)=0 \]

Step 3: Apply condition for real \(x\).
For real values of \(x\), the quadratic must have real roots.
Therefore, \[ D\geq 0 \] Here, \[ D=(-y-1)^2-4(y-1)(y-1) \] \[ =(y+1)^2-4(y-1)^2 \]

Step 4: Simplify the discriminant.
\[ (y+1)^2-4(y-1)^2\geq 0 \] \[ y^2+2y+1-4(y^2-2y+1)\geq 0 \] \[ y^2+2y+1-4y^2+8y-4\geq 0 \] \[ -3y^2+10y-3\geq 0 \] Multiplying by \(-1\), \[ 3y^2-10y+3\leq 0 \]

Step 5: Factorize.
\[ 3y^2-10y+3=0 \] \[ 3y^2-9y-y+3=0 \] \[ 3y(y-3)-1(y-3)=0 \] \[ (3y-1)(y-3)=0 \] So, \[ y=\frac{1}{3},\quad y=3 \] Since \[ 3y^2-10y+3\leq 0, \] we get \[ \frac{1}{3}\leq y\leq 3 \]

Step 6: Final conclusion.
Therefore, the range is \[ \boxed{\left[\frac{1}{3},3\right]} \]