Question

The sum of the maximum and minimum values of the function \[ f(x) = \frac{x^2 - x + 1}{x^2 + x + 1} \] is:

• A. \( \frac{17}{4} \)
• B. \( \frac{5}{2} \)
• C. \( \frac{10}{3} \)
• D. \( 0 \)

Hint:

To find extrema of rational functions, use the quotient rule and set \( f'(x) = 0 \). Evaluate at critical points to determine the function's range.

Answer 1

The Correct Option is C

Step 1: Differentiate \( f(x) \) 
We define: \[ f(x) = \frac{x^2 - x + 1}{x^2 + x + 1}. \] Using quotient rule, \[ f'(x) = \frac{(2x - 1)(x^2 + x + 1) - (x^2 - x + 1)(2x + 1)}{(x^2 + x + 1)^2}. \] Expanding the numerator: \[ (2x - 1)(x^2 + x + 1) = 2x^3 + 2x^2 + 2x - x^2 - x - 1 = 2x^3 + x^2 + x - 1. \] \[ (x^2 - x + 1)(2x + 1) = 2x^3 + x^2 - 2x^2 - x + 2x + 1 = 2x^3 - x^2 + x + 1. \] Thus, \[ f'(x) = \frac{(2x^3 + x^2 + x - 1) - (2x^3 - x^2 + x + 1)}{(x^2 + x + 1)^2}. \] \[ = \frac{2x^3 + x^2 + x - 1 - 2x^3 + x^2 - x - 1}{(x^2 + x + 1)^2}. \] \[ = \frac{2x^2 - 2}{(x^2 + x + 1)^2}. \] Setting \( f'(x) = 0 \), \[ 2(x^2 - 1) = 0. \] \[ x^2 = 1 \Rightarrow x = \pm 1. \] 

Step 2: Compute Maximum and Minimum Values 
\[ f(1) = \frac{1^2 - 1 + 1}{1^2 + 1 + 1} = \frac{1}{3}. \] \[ f(-1) = \frac{(-1)^2 - (-1) + 1}{(-1)^2 + (-1) + 1} = \frac{3}{2}. \] 

Step 3: Compute the sum 
\[ \frac{1}{3} + \frac{3}{2} = \frac{2}{6} + \frac{9}{6} = \frac{10}{6} = \frac{10}{3}. \] 

Step 4: Conclusion 
Thus, the correct answer is: \[ \mathbf{\frac{10}{3}.} \]