Question

The range of the function \( f(x) = \sqrt{3x^2 - 4x + 5} \) is: 
 

• A. \( (-\infty, \sqrt{\frac{11}{3}}) \)
• B. \( (-\infty, \sqrt{\frac{11}{5}}) \)
• C. \( \left[ \sqrt{\frac{11}{3}}, \infty \right) \)
• D. \( \left[ \sqrt{\frac{11}{5}}, \infty \right) \)

Hint:

To find the range of a function involving a square root, complete the square to rewrite the quadratic expression in a way that allows easy determination of the minimum value.

Answer 1

The Correct Option is C

The given function is: \[ f(x) = \sqrt{3x^2 - 4x + 5} \] Step 1: Completing the square to rewrite the quadratic expression. 
First, we complete the square for the quadratic expression \( 3x^2 - 4x + 5 \). 
Factor out the coefficient of \( x^2 \) from the first two terms: \[ f(x) = \sqrt{3(x^2 - \frac{4}{3}x) + 5} \] 
Now, complete the square inside the parentheses. 
The coefficient of \( x \) is \( \frac{-4}{3} \), so half of it is \( \frac{-2}{3} \), and squaring it gives \( \left( \frac{-2}{3} \right)^2 = \frac{4}{9} \). Add and subtract \( \frac{4}{9} \) inside the parentheses: \[ f(x) = \sqrt{3 \left( \left( x - \frac{2}{3} \right)^2 - \frac{4}{9} \right) + 5} \] Simplify: \[ f(x) = \sqrt{3 \left( x - \frac{2}{3} \right)^2 - \frac{4}{3} + 5} \] \[ f(x) = \sqrt{3 \left( x - \frac{2}{3} \right)^2 + \frac{11}{3}} \] 
Step 2: Determine the range. 
Since the expression inside the square root is always non-negative for all real values of \( x \), the minimum value of the function occurs when \( \left( x - \frac{2}{3} \right)^2 = 0 \), i.e., when \( x = \frac{2}{3} \). At \( x = \frac{2}{3} \), the value of \( f(x) \) is: \[ f\left( \frac{2}{3} \right) = \sqrt{\frac{11}{3}} \] Therefore, the range of the function is: \[ \left[ \sqrt{\frac{11}{3}}, \infty \right) \] Thus, the correct answer is (C).