3. The expression $$ \sqrt{-x^{2}+6 x-5} $$ has a (A) maximum value of 2 (B) minimum value of 2 (C) maximum value of 3 (D) minimum value of 3

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To determine the maximum or minimum value of the expression $$ \sqrt{-x^{2} + 6x - 5} $$, follow these steps:

1. **Determine the Domain:**
   The expression inside the square root must be non-negative:
   $$
   -x^{2} + 6x - 5 \geq 0
   $$
   This can be rewritten as:
   $$
   x^{2} - 6x + 5 \leq 0
   $$
   Factoring the quadratic equation $$ x^{2} - 6x + 5 = 0 $$ gives:
   $$
   (x - 1)(x - 5) = 0 \implies x = 1 \text{ or } x = 5
   $$
   Therefore, the domain is $$ 1 \leq x \leq 5 $$.

2. **Find the Maximum and Minimum of the Quadratic Expression:**
   Consider the quadratic function $$ y = -x^{2} + 6x - 5 $$.
   - This is a downward-opening parabola (since the coefficient of $$ x^2 $$ is negative).
   - The vertex (maximum point) occurs at:
     $$
     x = \frac{-b}{2a} = \frac{-6}{2(-1)} = 3
     $$
   - Substituting $$ x = 3 $$ into $$ y $$:
     $$
     y = -(3)^2 + 6(3) - 5 = -9 + 18 - 5 = 4
     $$
   - Thus, the maximum value of $$ y $$ is 4.

3. **Determine the Range of the Square Root:**
   - The square root function $$ \sqrt{y} $$ will range from $$ \sqrt{0} = 0 $$ to $$ \sqrt{4} = 2 $$.

4. **Conclusion:**
   The expression $$ \sqrt{-x^{2} + 6x - 5} $$ attains a **maximum value of 2** within its domain.

**Answer:**  
**(A) maximum value of 2**
The expression $$ \sqrt{-x^{2} + 6x - 5} $$ has a **maximum value of 2**. **Answer:** **(A) maximum value of 2**

3. The expression \( \sqrt{-x^{2}+6 x-5} \) has a (A) maximum value of 2 (B) minimum value of 2 (C) maximum value of 3 (D) minimum value of 3

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